Factorization Of Polynomials

(i) 3x²-4x+15 is a polynomial of one variable x.

(ii) y²+2_{is a polynomial of one variable y.}

(iii) 3+x is not a polynomial as the exponent of 3_{is not a positive integer.}

(iv) x - is not a polynomial as the exponent of - is not a positive integer.

(v) x¹²+y³+t⁵⁰ is a polynomial of three variables x, y, t.

Coefficient of x² in:

(i) 17 – 2x + 7x² is 7

(ii) 9-12x+x³ is 0

(iii) x² – 3x + 4 is

(iv) x-7 is 0

Degree of polynomial in:

(i) 7x³+4x²-3x+12 is 3

(ii) 12 -x+2x³ is 3

(iii) 5y- is 1

(iv) 7 is 0

(v) 0 is undefined

Given polynomial,

(i) x+x²+7y² is quadratic as degree of polynomial is 2.

(ii) 3x-2 is linear as degree of polynomial is 1.

(iii) 2x+x² is quadratic as degree of polynomial is 2.

(iv) 3y is linear as degree of polynomial is 1.

(v) t²+1 is quadratic as degree of polynomial is 2.

(vi) 7t⁴+4t³+3t-2 is bi-quadratic as degree of polynomial is 4.

(i) x²-xy+7y² is a polynomial in two variable x, y.

(ii) x²-2tx+7y²-x+t is a polynomial in two variable x, t.

(iii) t³-3t²+4t-5 is a polynomial in one variable t.

(iv) xy + yz + zx is a polynomial in three variable x, y, t.

(i) f(x) = 4x³-x²-3x+7 is a polynomial.

(ii) g(x) = 2x³-3x²+-1 is not a polynomial as exponent of x in is not a positive integer.

(iii) p(x) = x²-x+9 is a polynomial as all the exponents are positive integer.

(iv) q(x) = 2x²-3x++2 is not a polynomial as the exponent of x in is not a positive integer.

(v) h(x) = x⁴-+x-1 is not a polynomial as the exponent of x in –x^3/2 is not a positive integer.

(vi) f(x) = 2++4x is not a polynomial as the exponent of x in is not a positive integer.

Given polynomial,

(i) f(x) =0 is a constant polynomial as 0 is constant.

(ii) g(x) = 2x³-7x+4 is a cubic polynomial as degree of the polynomial is 3.

(iii) h(x) = -3x+ _{is a linear polynomial as the degree of polynomial is 1.}

(iv) p(x) = 2x²-x+4 is a quadratic polynomial as the degree of polynomial is 2.

(v) q(x) = 4x+3 is a linear polynomial as the degree of polynomial is 1.

(vi) r(x) = 3x³+4x²+5x-7 is a cubic polynomial as the degree of polynomial is 3.

Example of a binomial with degree 35 is 7x³⁵ – 5.

Example of a monomial with degree 100 is 2t¹⁰⁰.

We have,

f(x) = 2x³-13x²+17x+12

(i) f(2) = 2 (2)³ – 13 (2)² + 17 (2) + 12

= (2 * 8) – (13 * 4) + (17 * 2) + 12

= 16 – 52 + 34 + 12

= 10

(ii) f (-3) = 2 (-3)³ – 13 (-3)² + 17 (-3) + 12

= (2 * -27) – (13 * 9) + (17 * -3) + 12

= -54 – 117 – 51 + 12

= - 210

(iii) f (0) = 2 (0)³ – 13 (0)² + 17 (0) + 12

= 0 – 0 + 0 + 12

= 12

(i) f(x) = 3x + 1

_{Put x = -1/3}

_{f (-1/3) = 3 * (-1/3) + 1}

_{= -1 + 1}

_{= 0}

_{Therefore, x = -1/3 is a root of f (x) = 3x + 1}

_{(ii) We have,}

_{f (x) = x}² – 1

_{Put x = 1 and x = -1}

_{f (1) = (1)}² – 1 and f (-1) = (-1)² – 1

_{= 1 – 1 = 1- 1}

_{= 0 = 0}

Therefore, x = -1 and x = 1 are the roots of f(x) = x² – 1

(iii) g (x) = 3x² – 2

Put x = and x =

g () = 3 ()² – 2 and g () = 3 ()² – 2

= 3 * – 2 = 3 * – 2

= 2 0 = 2 0

Therefore, x = and x = are not the roots of g (x) = 3x² – 2

(iv) p (x) = x³ – 6x² + 11x – 6

Put x = 1

p (1) = (1)³ – 6 (1)² + 11 (1) – 6

= 1 – 6 + 11 – 6

= 0

Put x = 2

p (2) = (2)³ – 6 (2)² + 11 (2) – 6

= 8 – 24 + 22 – 6

= 0

Put x = 3

p (3) = (3)³ – 6 (3)² + 11 (3) – 6

= 27 – 54 + 33 – 6

= 0

Therefore, x = 1, 2, 3 are roots of p (x) = x³ – 6x² + 11x – 6

(v) f (x) = 5x –

Put x =

f () = 5 * –

= 4 – 0

Therefore, x = is not a root of f (x) = 5x –

(vi) f (x) = x²

Put x = 0

f (0) = (0)²

= 0

Therefore, x = 0 is not a root of f (x) = x²

(vii) f (x) = lx + m

Put x =

f () = l * () + m

= -m + m

= 0

Therefore, x = is a root of f (x) = lx + m

(viii) f (x) = 2x + 1

Put x =

f () = 2 * + 1

= 1 + 1

= 2 0

Therefore, x = is not a root of f (x) = 2x + 1

We have,

f (x) = 2x² – 3x + 7a

Put x = 2

f (2) = 2 (2)² – 3 (2) + 7a

= 2 * 4 – 6 + 7a

= 8 – 6 + 7a

= 2 + 7a

Given, x = 2 is a root of f (x) = 2x² – 3x + 7a

f (2) = 0

Therefore, 2 + 7a = 0

7a = -2

a =

We have,

p (x) = 8x³ - ax² - x + 2

Put x =

p () = 8 ()³ – a ()² – () + 2

= 8 × – a × + + 2

= -1 - + + 2

= -

Given that,

x = is a root of p (x)

p () = 0

Therefore,

- = 0

=

2a = 12

a = 6

we have,

f (x) = 2x³-3x²+ax+b

Put,

x = 0

f (0) = 2 (0)³ – 3 (0)² + a (0) + b

= 0 – 0 + 0 + b

= b

x = -1

f (-1) = 2 (-1)³ – 3 (-1)² + a (-1) + b

= -2 – 3 – a + b

= -5 – a + b

Since, x = 0 and x = -1 are roots of f (x)

f (0) = 0 and f (-1) = 0

b = 0 and -5 – a + b = 0

= a – b = -5

= a – 0 = -5

= a = -5

Therefore, a = -5 and b = 0

We have,

f(x) = x³+6x²+11x+6

Clearly, f (x) is a polynomial with integer coefficient and the coefficient of the highest degree term i.e., the leading coefficient is 1.

Therefore, integer root of f (x) are limited to the integer factors of 6, which are:

We observe that

f (-1) = (-1)³ + 6 (-1)² + 11 (-1) + 6

= -1 + 6 -11 + 6

= 0

f (-2) = (-2)³ + 6 (-2)² + 11 (-2) + 6

= -8 + 24 – 22 + 6

= 0

f (-3) = (-3)³ + 6 (-3)² + 11 (-3) + 6

= -27 + 54 – 33 + 6

= 0

Therefore, integral roots of f (x) are -1, -2, -3.

We have,

f(x) = 2x³+x²-7x-6

Clearly, f (x) is a cubic polynomial with integer coefficients. If is a rational root in lowest term, then the value of b are limited to the factors of 6 which are and values of c are limited to the factors of 2 which are .

Hence, the possible rational roots of f(x) are:

We observe that,

f (-1) = 2 (-1)³ + (-1)² – 7 (-1) – 6

= -2 + 1 + 7 – 6

= 0

f (2) = 2 (2)³ + (2)² – 7 (2) – 6

= 16 + 4 – 14 – 6

= 0

f () = 2 ()³ + ()² – 7 () – 6

= + + – 6

= 0

Hence, -1, 2, are the rational roots of f (x).

We have,

f(x) = x³+4x²-3x+10 and g (x) = x + 4

Therefore, by remainder theorem when f (x) is divided by g (x) = x – (-4), the remainder is equal to f (-4)

Now, f(x) = x³+4x²-3x+10

f (-4) = (-4)³ + 4 (-4)² – 3 (-4) + 10

= -64 + 4 * 16 + 12 + 10

= 22

Hence, required remainder is 22.

We have,

f(x) = 4x⁴-3x³-2x²+x-7 and g(x) = x-1

Therefore, by remainder theorem when f (x) is divided by g (x) = x – 1, the remainder is equal to f (+1)

Now, f(x) = 4x⁴-3x³-2x²+x-7

f (1) = 4 (1)⁴ – 3 (1)³ – 2 (1)² + 1 – 7

= 4 – 3 – 2 + 1 – 7

= -7

Hence, required remainder is -7.

We have,

f(x) = 2x⁴-6x³+2x²-x+2 and g(x) = x+2

Therefore, by remainder theorem when f (x) is divided by g (x) = x – (-2), the remainder is equal to f (-2)

Now, f(x) = 2x⁴-6x³+2x²-x+2

f (-2) = 2 (-2)⁴ – 6 (-2)³ + 2 (-2)² – (-2) + 2

= 2 * 16 + 48 + 8 + 2 + 2

= 32 + 48 + 12

= 92

Hence, required remainder is 92.

We have,

f(x) = 4x³-12x²+14x-3 and g(x) = 2x-1

Therefore, by remainder theorem when f (x) is divided by g (x) = 2 (x - ), the remainder is equal to f ()

Now, f(x) = 4x³-12x²+14x-3

f () = 4 ()³ – 12 ()² + 14 () – 3

= (4 * ) – (12 * ) + 7 – 3

= – 3 + 7 – 3

=

Hence, required remainder is

We have,

f(x) = x³-6x²+2x-4 and g(x) = 1-2x

Therefore, by remainder theorem when f (x) is divided by g (x) = -2 (x - ), the remainder is equal to f ()

Now, f(x) = x³-6x²+2x-4

f () = ()³ – 6 ()² + 2 () – 4

= - + 1 – 4

=

Hence, required remainder is

We have,

f(x) = x⁴-3x²+4 and g(x) = x-2

Therefore, by remainder theorem when f (x) is divided by g (x) = x – 2, the remainder is equal to f (2)

Now, f(x) = x⁴-3x²+4

f (2) = (2)⁴ – 3 (2)² + 4

= 16 – 12 + 4

= 8

Hence, required remainder is 8.

We have,

f(x) = 9x³-3x²+x-5 and g(x) = x=

Therefore, by remainder theorem when f (x) is divided by g (x) = x - , the remainder is equal to f ()

Now, f(x) = 9x³-3x²+x-5

f () = 9 ()³ – 3 ()² + – 5

= (9 * ) – (3 * ) + – 5

= - + – 5

= 2 – 5 = -3

Hence, the required remainder is -3.

We have,

f(x) = 3x⁴+2x³ and g(x) = x+

Therefore, by remainder theorem when f (x) is divided by g (x) = x – (- ), the remainder is equal to f ()

Now, f(x) = 3x⁴+2x³

f () = 3 ()⁴ + 2 ()³ – () - +

= 3 * + 2 * - - +

= - - + +

= =

= 0

Hence, required remainder is 0.

Let, p (x) = 2x³+ax²+3x-5 and q (x) = x³+x²-4x+a be the given polynomials.

The remainders when p (x) and q (x) are divided by (x – 2) and p (2) and q (2) respectively.

By the given condition, we have:

p (2) = q (2)

2 (2)³ + a (2)² + 3 (2) – 5 = (2)³ + (2)² – 4 (2) + a

16 + 4a + 6 – 5 = 8 + 4 – 8 + a

3a + 13 = 0

3a = - 13

a =

Let, p (x) = ax³+3x²-3 and q (x) = 2x³-5x+a be the given polynomials.

Now,

R₁ = Remainder when p (x) is divided by (x – 4)

= p (4)

= a (4)³ + 3 (4)² – 3 [Therefore, p (x) = ax³+3x²-3]

= 64a + 48 – 3

R₁ = 64a + 45

And,

R₂ = Remainder when q (x) is divided by (x – 4)

= q (4)

= 2 (4)³ – 5 (4) + a [Therefore, q (x) = 2x³-5x+a]

= 128 – 20 + a

R₂ = 108 + a

(i) Given condition is,

R₁ = R₂

64a + 45 = 108 + a

63a – 63 = 0

63a = 63

a = 1

(ii) Given condition is R₁ + R₂ = 0

64a + 45 + 108 + a = 0

65a + 153 = 0

65a = -153

a =

(iii) Given condition is 2R₁ – R₂ = 0

2 (64a + 45) – (108 + a) = 0

128a + 90 – 108 – a

127a – 18 = 0

127a = 18

a =

Let p (x) = ax³+3x²-13 and q (x) = 2x³-5x+a be the given polynomials.

The remainders when p (x) and q (x) are divided by (x – 2) and p (2) and q (2) respectively.

By the given condition, we have:

p (2) = q (2)

a (2)³ + 3 (2)² – 13 = 2 (2)³ – 5 (2) + a

8a + 12 – 13 = 16 – 10 + a

7a – 7 = 0

7a = 7

a =

= 1

Let, f(x) = x³+3x²+3x+1

(i) x + 1

Apply remainder theorem

⇒ x + 1 =0

⇒ x = - 1

Replace x by – 1 we get

⇒ x³+3x² + 3x + 1

⇒ (-1)³ + 3(-1)² + 3(-1) + 1

⇒ -1 + 3 - 3 + 1

⇒ 0

Hence, the required remainder is 0.

(ii)x-

Apply remainder theorem

⇒ x – 1/2 =0

⇒ x = 1/2

Replace x by 1/2 we get

⇒ x³+3x² + 3x + 1

⇒ (1/2)³ + 3(1/2)² + 3(1/2) + 1

⇒ 1/8 + 3/4 + 3/2 + 1

Add the fraction taking LCM of denominator we get

⇒ (1 + 6 + 12 + 8)/8

⇒ 27/8

Hence, the required remainder is 27/8

(iii) x = x – 0

By remainder theorem required remainder is equal to f (0)

Now, f (x) = x³+3x²+3x+1

f (0) = (0)³ + 3 (0)² + 3 (0) + 1

= 0 + 0 + 0 + 1

= 1

Hence, the required remainder is 1.

(iv) x+π = x – (-π)

By remainder theorem required remainder is equal to f (-π)

Now, f (x) = x³ + 3x² + 3x + 1

f (- π) = (- π)³ + 3 (- π)² + 3 (- π) + 1

= - π³ + 3π² - 3π + 1

Hence, required remainder is - π³ + 3π² - 3π + 1.

(v) 5 + 2x = 2 [x – ()]

By remainder theorem required remainder is equal to f ()

Now, f (x) = x³ + 3x² + 3x + 1

f () = )³ + 3 ()² + 3 () + 1

= + 3 * + 3 * + 1

= + - + 1

=

Hence, the required remainder is .

We have,

f(x) = x³-6x²+11x-6 and g(x) = x-3

In order to find whether polynomials g (x) = x – 3 is a factor of f (x), it is sufficient to show that f (3) = 0

Now,

f(x) = x³-6x²+11x-6

f (3) = 3³ – 6 (3)² + 11 (3) – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = 3x⁴+17x³+9x²-7x-10 and g(x) = x+5

In order to find whether the polynomials g (x) = x – (-5) is a factor of f (x) or not, it is sufficient to show that f (-5) = 0

Now,

f(x) = 3x⁴+17x³+9x²-7x-10

f (-5) = 3 (-5)⁴ + 17 (-5)³ + 9 (-5)² – 7 (-5) – 10

= 3 * 625 + 17 * (-125) + 9 * 25 + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = x⁵+3x⁴-x³-3x²+5x+15 and g(x) = x+3

In order to find whether g (x) = x – (-3) is a factor of f (x) or not, it is sufficient to prove that f (-3) = 0

Now,

f(x) = x⁵+3x⁴-x³-3x²+5x+15

f (-3) = (-3)⁵ + 3 (-3)⁴ – (-3)³ – 3 (-3)² + 5 (-3) + 15

= - 243 + 243 – (-27) – 3 (9) + 5 (-3) + 15

= - 243 + 243 + 27 – 27 – 15 + 15

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = x³-6x²-19x+84 and g(x) = x-7

In order to find whether g (x) = x – 7 is a factor of f (x) or not, it is sufficient to show that f (7) = 0

Now,

f(x) = x³-6x²-19x+84

f (7) = (7)³ – 6 (7)² – 19 (7) + 84

= 343 – 294 – 133 + 84

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = 3x³+x²-20x+12 and g(x) = 3x-2

In order to find whether g (x) is = 3x – 2 is a factor of f (x) or not, it is sufficient to show that f () = 0

Now,

f(x) = 3x³+x²-20x+12

f () = 3 ()³ + ()² – 20 () + 12

= - + 12

=

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = 2x³-9x²+x+12 and g(x) = 3-2x

In order to find g (x) = 3 – 2x = 2 (x - ) is a factor of f (x) or not, it is sufficient to prove that f () = 0

Now,

f(x) = 2x³-9x²+x+12

f () = 2 ()³ – 9 ()² + + 12

= - + + 12

=

= 0

Hence, g (x) is a factor of f (x).

We have,

f(x) = x³-6x²+11x-6 and g(x) = x²-3x+2

In order to find g (x) = x²-3x+2 = (x – 1) (x – 2) is a factor of f (x) or not, it is sufficient to prove that (x – 1) and (x – 2) are factors of f (x)

i.e. We have to prove that f (1) = 0 and f (2) = 0

f (1) = (1)³ – 6 (1)² + 11 (1) – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

f (2) = (2)³ – 6 (2)² + 11 (2) – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

Since, (x – 1) and (x – 2) are factors of f (x).

Therefore, g (x) = (x – 1) (x – 2) are the factors of f (x).

Let, f (x) = x³-3x²-10x+24 be the given polynomial.

In order to prove that (x – 2) (x + 3) (x – 4) are the factors of f (x), it is sufficient to show that f (2) = 0, f (-3) = 0 and f (4) = 0 respectively.

Now,

f (x) = x³-3x²-10x+24

f (2) = (2)³ – 3 (2)² – 10 (2) + 24

= 8 – 12 – 20 + 24

= 0

f (-3) = (-3)³ – 3 (-3)² – 10 (-3) + 24

= -27 – 27 + 30 + 24

= 0

f (4) = (4)³ – 3 (4)² – 10 (4) + 24

= 64 – 48 – 40 + 24

= 0

Hence, (x – 2), (x + 3) and (x – 4) are the factors of the given polynomial.

Let f (x) = x³-6x²-19x+84 be the given polynomial.

In order to prove that (x + 4), (x – 3) and (x – 7) are factors of f (x), it is sufficient to prove that f (-4) = 0, f (3) = 0 and f (7) = 0 respectively.

Now,

f (x) = x³-6x²-19x+84

f (-4) = (-4)³ – 6 (-4)² – 19 (-4) + 84

= -64 – 96 + 76 + 84

= 0

f (3) = (3)³ – 6 (3)² – 19 (3) + 84

= 27 – 54 – 57 + 84

= 0

f (7) = (7)³ – 6 (7)² – 19 (7) + 84

= 343 – 294 – 133 + 84

= 0

Hence, (x – 4), (x – 3) and (x -7) are the factors of the given polynomial x³-6x²-19x+84.

Let, f (x) = x³-3x²+ax-10 be the given polynomial.

By factor theorem,

If (x – 5) is a factor of f (x) then f (5) = 0

Now,

f (x) = x³-3x²+ax-10

f (5) = (5)³ – 3 (5)² + a (5) – 10

0 = 125 – 75 + 5a – 10

0 = 5a + 40

a = -8

Hence, (x – 5) is a factor of f (x), if a = - 8.

Let f(x) = 5x³-7x²-ax-28 be the given polynomial.

From factor theorem,

If (x -4) is a factor of f (x) then f (4) = 0

f (4) = 0

0 = 5 (4)³ – 7 (4)² – a (4) – 28

0 = 320 – 112 – 4a – 28

0 = 180 – 4a

4a = 180

a = 45

Hence, (x – 4) is a factor of f (x) when a = 45.

Let, f (x) = 4x⁴+2x³-3x²+8x+5a

f (-2) = 0

4 (-2)⁴ + 2 (-2)³ – 3 (-2)² + 8 (-2) + 5a = 0

64 – 16 – 12 – 16 + 5a = 0

5a = - 20

a = -4

Hence, (x + 2) is a factor f (x) when a = -4.

Let, f (x) = k²x³ - kx² + 3kx - k

By factor theorem,

If (x – 3) is a factor of f (x) then f (3) = 0

k² (3)³ – k (3)² + 3 k (3) – k = 0

27k² – 9k + 9k – k = 0

k (27k – 1) = 0

k = 0 or (27k – 1) = 0

k = 0 or k =

Hence, (x – 3) is a factor of f (x) when k = 0 or k = .

Let, f (x) = ax⁴+2x³-3x²+bx-4 and g (x) = x ² – 4

We have,

g (x) = x² – 4

= (x – 2) (x + 2)

Given,

g (x) is a factor of f (x)

(x – 2) and (x + 2) are factors of f (x).

From factor theorem if (x – 2) and (x + 2) are factors of f (x) then f (2) = 0 and f (-2) = 0 respectively.

f (2) = 0

a * (-2)⁴ + 2 (2)³ – 3 (2)² + b (2) – 4 = 0

16a – 16 – 12 + 2b – 4 = 0

16a + 2b = 0

2 (8a + b) = 0

8a + b = 0 (i)

Similarly,

f (-2) = 0

a * (-2)⁴ + 2 (-2)³ – 3 (-2)² + b (-2) – 4 = 0

16a – 16 – 12 - 2b – 4 = 0

16a - 2b – 32 = 0

16a – 2b – 32 = 0

2 (8a - b) = 32

8a – b = 16 (ii)

Adding (i) and (ii), we get

8a + b + 8a – b = 16

16a = 16

a = 1

Put a = 1 in (i), we get

8 * 1 + b = 0

b = -8

Hence, a = 1 and b = -8.

Let, f (x) = x³+3x²-2αx+β be the given polynomial,

From factor theorem,

If (x + 1) and (x + 2) are factors of f (x) then f (-1) = 0 and f (-2) = 0

f (-1) = 0

(-1)³ + 3 (-1)² – 2 α (-1) + β = 0

-1 + 3 + 2 α + β = 0

2 α + β + 2 = 0 (i)

Similarly,

f (-2) = 0

(-2)³ + 3 (-2)² – 2 α (-2) + β = 0

-8 + 12 + 4 α + β = 0

4 α + β + 4 = 0 (ii)

Subtract (i) from (ii), we get

4 α + β + 4 – (2 α + β + 2) = 0 – 0

4 α + β + 4 - 2 α - β - 2 = 0

2 α + 2 = 0

α = -1

Put α = -1 in (i), we get

2 (-1) + β + 2 = 0

β = 0

Hence, α = -1 and β = 0.

Let, f (x) = x⁴+px³+2x²-3x+q be the given polynomial.

And, let g (x) = (x² – 1) = (x – 1) (x + 1)

Clearly,

(x – 1) and (x + 1) are factors of g (x)

Given, g (x) is a factor of f (x)

(x – 1) and (x + 1) are factors of f (x)

From factor theorem

If (x – 1) and (x + 1) are factors of f (x) then f (1) = 0 and f (-1) = 0 respectively.

f (1) = 0

(1)⁴ + p (1)³ + 2 (1)² – 3 (1) + q = 0

1 + p + 2 – 3 + q = 0

p + q = 0 (i)

Similarly,

f (-1) = 0

(-1)⁴ + p (-1)³ + 2 (-1)² - 3 (-1) + q = 0

1 – p + 2 + 3 + q = 0

q – p + 6 = 0 (ii)

Adding (i) and (ii), we get

p + q + q – p + 6 = 0

2q + 6 = 0

2q = - 6

q = -3

Putting value of q in (i), we get

p – 3 = 0

p = 3

Hence, x² – 1 is divisible by f (x) when p = 3 and q = - 3.

Let, f (x) = x⁴+ax³-3x²+2x+b be the given polynomial

From factor theorem

If (x + 1) and (x – 1) are factors of f (x) then f (-1) = 0 and f (1) = 0 respectively.

f (-1) = 0

(-1)⁴ + a (-1)³ – 3 (-1)² + 2 (-1) + b = 0

1 – a – 3 – 2 + b = 0

b – a – 4 = 0 (i)

Similarly, f (1) = 0

(1)⁴ + a (1)³ – 3 (1)² + 2 (1) + b = 0

1 + a – 3 + 2 + b = 0

a + b = 0 (ii)

Adding (i) and (ii), we get

2b – 4 = 0

2b = 4

b = 2

Putting the value of b in (i), we get

2 – a – 4 = 0

a = -2

Hence, a = -2 and b = 2.

Let f (x) = x³+ax²-bx+10 and g (x) = x²-3x+2 be the given polynomials.

We have g (x) = x²-3x+2 = (x – 2) (x – 1)

Clearly, (x -1) and (x – 2) are factors of g (x)

Given that f (x) is divisible by g (x)

g (x) is a factor of f (x)

(x – 2) and (x – 1) are factors of f (x)

From factor theorem,

If (x – 1) and (x – 2) are factors of f (x) then f (1) = 0 and f (2) = 0 respectively.

f (1) = 0

(1)³ + a (1)² – b (1) + 10 = 0

1 + a – b + 10 = 0

a – b + 11 = 0 (i)

f (2) = 0

(2)³ + a (2)² - b (2) + 10 = 0

8 + 4a – 2b + 10 = 0

4a – 2b + 18 = 0

2 (2a – b + 9) = 0

2a – b + 9 = 0 (ii)

Subtract (i) from (ii), we get

2a – b + 9 – (a – b + 11) = 0

2a – b + 9 – a + b – 11 = 0

a – 2 = 0

a = 2

Putting value of a in (i), we get

2 – b + 11 = 0

b = 13

Hence, a = 2 and b = 13

Let, f (X) = ax³+x²-2x+b be the given polynomial.

Given (x + 1) and (x – 1) are factors of f (x).

From factor theorem,

If (x + 1) and (x – 1) are factors of f (x) then f (-1) = 0 and f (1) = 0 respectively.

f (-1) = 0

a (-1)³ + (-1)² – 2 (-1) + b = 0

-a + 1 + 2 + b = 0

-a + 3 + b = 0

b – a + 3 = 0 (i)

f (1) = 0

a (1)³ + (1)² – 2 (1) + b = 0

a + 1 – 2 + b = 0

a + b – 1 = 0

b + a – 1 = 0 (ii)

Adding (i) and (ii), we get

b – a + 3 + b + a – 1 = 0

2b + 2 = 0

2b = - 2

b = -1

Putting value of b in (i), we get

-1 - a + 3 = 0

-a + 2 = 0

a = 2

Hence, the value of a = 2 and b = -1.

Let p (x) = x³-3x²-12x+19 and q (x) = x²+x-6

By division algorithm, when p (x) is divided by q (x), the remainder is a linear expression in x.

So, let r (x) = ax + b is added to p (x) so that p (x) + r (x) is divisible by q (x).

Let,

f (x) = p (x) + r (x)

= x³ - 3x² - 12x + 19 + ax + b

= x³ – 3x² + x (a – 12) + b + 19

We have,

q (x) = x²+x-6

= (x + 3) (x – 2)

Clearly, q (x) is divisible by (x – 2) and (x + 3) i.e. (x – 2) and (x + 3) are factors of q (x)

We have,

f (x) is divisible by q (x)

(x – 2) and (x + 3) are factors of f (x)

From factor theorem,

If (x – 2) and (x + 3) are factors of f (x) then f (2) = 0 and f (-3) = 0 respectively.

f (2) = 0

(2)³ – 3 (2)² + 2 (a – 12) + b + 19 = 0

⇒ 8 – 12 + 2a – 24 + b + 19 = 0

⇒ 2a + b – 9 = 0 (i)

Similarly,

f (-3) = 0

(-3)³ – 3 (-3)² + (-3) (a – 12) + b + 19 = 0

⇒ -27 – 27 – 3a + 36 + b + 19 = 0

⇒ b – 3a + 1 = 0 (ii)

Subtract (i) from (ii), we get

b – 3a + 1 – (2a + b – 9) = 0 – 0

⇒ b – 3a + 1 – 2a – b + 9 = 0

⇒ - 5a + 10 = 0

⇒ 5a = 10

⇒ a = 2

Put a = 2 in (ii), we get

b – 3 × 2 + 1 = 0

⇒ b – 6 + 1 = 0

⇒ b – 5 = 0

⇒ b = 5

Therefore, r (x) = ax + b

= 2x + 5

Hence, x³ – 3x – 12x + 19 is divisible by x² + x – 6 when 2x + 5 is added to it.

Let p (x) = x³ - 6x² - 15x + 80 and q (x) = x² + x - 12

By division algorithm, when p (x) is divided by q (x), the remainder is a linear expression in x.

So, let r (x) = ax + b is subtracted to p (x) so that p (x) + r (x) is divisible by q (x).

Let, f (x) = p (x) – r (x)

⇒ f(x) = x³ - 6x² - 15x + 80 – (ax + b)

⇒ f(x) = x³ - 6x² – (a + 15)x + (80 – b)

We have,

q(x) = x² + x – 12

⇒ q(x) = (x + 4) (x - 3)

Clearly, q (x) is divisible by (x + 4) and (x - 3) i.e. (x + 4) and (x - 3) are factors of q (x)

Therefore, f (x) will be divisible by q (x), if (x + 4) and (x - 3) are factors of f (x).

i.e. f(-4) = 0 and f(3) = 0

f (3) = 0

⇒ (3)³ – 6(3)² – 3 (a + 15) + 80 – b = 0

⇒ 27 – 54 – 3a – 45 + 80 – b = 0

⇒ 8 – 3a – b = 0 (i)

f (-4) = 0

⇒ (-4)³ – 6 (-4)² – (-4) (a + 15) + 80 – b = 0

⇒ -64 – 96 + 4a + 60 + 80 – b = 0

⇒ 4a – b – 20 = 0 (ii)

Subtract (i) from (ii), we get

⇒ 4a – b – 20 – (8 – 3a – b) = 0

⇒ 4a – b – 20 – 8 + 3a + b = 0

⇒ 7a = 28

⇒ a = 4

Put value of a in (ii), we get

⇒ b = -4

Putting the value of a and b in r (x) = ax + b, we get

r (x) = 4x – 4

Hence, p (x) is divisible by q (x), if r (x) = 4x – 4 is subtracted from it.

Let p (x) = 3x³ + x² - 22x + 9 and q (x) = 3x² + 7x - 6.

By division algorithm,

When p (x) is divided by q (x), the remainder is a linear expression in x.

So, let r (x) = ax + b is added to p (x) so that p (x) + r (x) is divisible by q (x).

Let, f (x) = p (x) + r (x)

= 3x³ + x² – 22x + 9 + (ax + b)

= 3x³ + x² + x (a – 22) + b + 9

We have,

q (x) = 3x² + 7x – 6

q (x) = 3x (x + 3) – 2 (x + 3)

q (x) = (3x – 2) (x + 3)

Clearly, q (x) is divisible by (3x – 2) and (x + 3). i.e. (3x – 2) and (x + 3) are factors of q(x),

Therefore, f(x) will be divisible by q(x), if (3x – 2) and (x + 3) are factors of f(x).

i.e. f (2/3) = 0 and f (-3) = 0 [∵ 3x – 2 = 0, x = 2/3 and x + 3 = 0, x = -3]

f (2/3) = 0

⇒ 6a + 9b – 39 = 0

⇒ 3 (2a + 3b – 13) = 0

⇒ 2a + 3b – 13 = 0 (i)

Similarly,

f (-3) = 0

⇒ 3 (-3)³ + (-3)² + (-3) (a – 2x) + b + 9 = 0

⇒ -81 + 9 – 3a + 66 + b + 9 = 0

⇒ b – 3a + 3 = 0

⇒ 3 (b – 3a + 3) = 0

⇒ 3b – 9a + 9 = 0 (ii)

Subtract (i) from (ii), we get

3b – 9a + 9 – (2a + 3b – 13) = 0

3b – 9a + 9 – 2a – 3b + 13 = 0

⇒ -11a + 22 = 0

⇒ a = 2

Putting value of a in (i), we get

⇒ b = 3

Putting the values of a and b in r (x) = ax + b, we get

r (x) = 2x + 3

Hence, p (x) is divisible by q (x) if r (x) = 2x + 3 is divisible by it.

(i) Let, f (x) = x³-2ax²+ax-1 be the given polynomial

From factor theorem,

If (x – 2) is a factor of f (x) then f (2) = 0 [Therefore, x – 2 = 0, x = 2]

f (2) = 0

(2)³ – 2 a (2)² + a (2) – 1 = 0

8 – 8a + 2a – 1 = 0

7 – 6a = 0

6a = 7

a =

Hence, (x – 2) is a factor of f (x) when a = .

(ii) Let f (x) = x⁵-3x⁴-ax³+3ax²+2ax+4 be the given polynomial

From factor theorem,

If (x – 2) is a factor of f (x) then f (2) = 0 [Therefore, x – 2= 0, x = 2]

f (2) = 0

(2)⁵ – 3 (2)⁴ – a (2)³ + 3 a (2)² + 2 a (2) + 4 = 0

32 – 48 – 8a + 12a + 4a + 4 = 0

-12 + 8a = 0

8a = 12

a =

Hence, (x – 2) is a factor of f (x) when a = .

(i) Let f (x) = x⁶-ax⁵+x⁴-ax³+3x-a+2 be the given polynomial

From factor theorem,

If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a]

f (a) = 0

(a)⁶ – a (a)⁵ + (a)⁴ – a (a)³ + 3 (a) – a + 2 = 0

a⁶ – a⁶ + a⁴ – a⁴ + 3a – a + 2 = 0

2a + 2 = 0

a = -1

Hence, (x – a) is a factor f (x) when a = -1.

(ii) Let, f (x) = x⁵-a²x³+2x+a+1 be the given polynomial

From factor theorem,

If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a]

f (a) = 0

(a)⁵ – a² (a)³ + 2 (a) + a + 1 = 0

a⁵ – a⁵ + 2a + a + 1 = 0

3a + 1 = 0

3a = -1

a =

Hence, (x – a) is a factor f (x) when a = .

(i) Let, f (x) = x³+ax²-2x+a+4 be the given polynomial

From factor theorem,

If (x + a) is a factor of f (x) then f (-a) = 0 [Therefore, x + a = 0, x = -a]

f (-a) = 0

(-a)³ + a (-a)² - 2 (-a) + a + 4 = 0

- a³ + a³ + 2a + a + 4 = 0

3a + 4 = 0

3a = -4

a =

Hence, (x + a) is a factor f (x) when a = .

(ii) Let, f (x) = x⁴-a²x²+3x-a be the given polynomial

From factor theorem,

If (x + a) is a factor of f (x) then f (-a) = 0 [Therefore, x + a = 0, x = -a]

f (-a) = 0

(-a)⁴ – a² (-a)² + 3 (-a) - a = 0

a⁴ – a⁴ - 3a - a = 0

-4a = 0

a = 0

Hence, (x + a) is a factor f (x) when a = 0.

Let f (x) = x³+6x²+11x+6 be the given polynomial.

The constant term in f (x) is 6 and factors of 6 are

Putting x = - 1 in f (x) we have,

f (-1) = (-1)³ + 6 (-1)² + 11 (-1) + 6

= -1 + 6 – 11 + 6

= 0

Therefore, (x + 1) is a factor of f (x)

Similarly, (x + 2) and (x + 3) are factors of f (x).

Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (x) = k (x + 1) (x + 2) (x + 3)

x³+6x²+11x+6 = k (x + 1) (x + 2) (x + 3)

Putting x = 0, on both sides we get,

0 + 0 + 0 + 6 = k (0 + 1) (0 + 2) (0 + 3)

6 = 6k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x + 3), we get

f (x) = (x + 1) (x + 2) (x + 3)

Hence,

x³+6x²+11x+6 = (x + 1) (x + 2) (x + 3)

Let, f (x) = x³+2x²-x-2

The constant term in f (x) is equal to -2 and factors of -2 are .

Putting x = 1 in f (x), we have

f (1) = (1)³ + 2 (1)² – 1 – 2

= 1 + 2 – 1 – 2

= 0

Therefore, (x – 1) is a factor of f (x).

Similarly, (x + 1) and (x + 2) are the factors of f (x).

Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (x) = k (x – 1) (x + 1) (x + 2)

x³+2x²-x-2 = k (x – 1) (x + 1) (x + 2)

Putting x = 0 on both sides, we get

0 + 0 – 0 – 2 = k (0 – 1) (0 + 1) (0 + 2)

-2 = -2k

k = 1

Putting k = 1 in f (x) = k (x – 1) (x + 1) (x + 2), we get

f (x) = (x – 1) (x + 1) (x + 2)

Hence,

x³+2x²-x-2 = (x – 1) (x + 1) (x + 2)

Let, f (x) = x³-6x²+3x+10

The constant term in f (x) is equal to 10 and factors of 10 are ,

Putting x = - 1 in f (x), we have

f (-1) = (-1)³ – 6 (-1)² + 3 (-1) + 10

= -1 – 6 – 3 + 10

= 0

Therefore, (x + 1) is a factor of f (x).

Similarly, (x - 2) and (x - 5) are the factors of f (x).

Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (x) = k (x + 1) (x - 2) (x - 5)

x³-6x²+3x+10 = k (x + 1) (x - 2) (x - 5)

Putting x = 0 on both sides, we get

0 + 0 – 0 + 10 = k (0 + 1) (0 - 2) (0 - 5)

10 = 10k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x - 2) (x - 5), we get

f (x) = (x + 1) (x - 2) (x - 5)

Hence,

x³-6x²+3x+10 = (x + 1) (x - 2) (x - 5)

Let, f (x) = x⁴-7x³+9x²+7x-10

The constant term in f (x) is equal to -10 and factors of -10 are ,

Putting x = 1 in f (x), we have

f (1) = (1)⁴ – 7 (1)³ + 9 (1)² + 7 (1) - 10

= 1 – 7 + 9 + 7 - 10

= 0

Therefore, (x - 1) is a factor of f (x).

Similarly, (x + 1), (x - 2) and (x - 5) are the factors of f (x).

Since, f (x) is a polynomial of degree 4. So, it cannot have more than four linear factors.

Therefore, f (x) = k (x – 1) (x + 1) (x - 2) (x - 5)

x⁴-7x³+9x²+7x-10 = k (x – 1) (x + 1) (x - 2) (x - 5)

Putting x = 0 on both sides, we get

0 + 0 – 0 - 10 = k (0 – 1) (0 + 1) (0 - 2) (0 - 5)

-10 = -10k

k = 1

Putting k = 1 in f (x) = k (x – 1) (x + 1) (x - 2) (x - 5), we get

f (x) = (x – 1) (x + 1) (x - 2) (x - 5)

Hence,

x⁴-7x³+9x²+7x-10 = (x – 1) (x + 1) (x - 2) (x - 5)

Let, f (x) = x⁴-2x³-7x²+8x+12

The constant term in f (x) is equal to +12 and factors of +12 are ,

Putting x = - 1 in f (x), we have

f (-1) = (-1)⁴ – 2 (-1)³ – 7 (-1)² + 8 (-1) + 12

= 1 + 2 – 7 – 8 + 12

= 0

Therefore, (x + 1) is a factor of f (x).

Similarly, (x + 2), (x – 2) and (x - 3) are the factors of f (x).

Since, f (x) is a polynomial of degree 4. So, it cannot have more than four linear factors.

Therefore, f (x) = k (x + 1) (x + 2) (x - 2) (x - 3)

x⁴-2x³-7x²+8x+12 = k (x + 1) (x + 2) (x - 2) (x - 3)

Putting x = 0 on both sides, we get

0 - 0 – 0 + 0 + 12 = k (0 + 1) (0 + 2) (0 - 2) (0 - 3)

12 = 12k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x - 2) (x - 3), we get

f (x) = (x + 1) (x + 2) (x - 2) (x - 3)

Hence,

x⁴-2x³-7x²+8x+12 = (x + 1) (x + 2) (x - 2) (x - 3)

Let, f (x) = x⁴+10x³+35x²+50x +24

The constant term in f (x) is equal to +24 and factors of +24 are ,

Putting x = - 1 in f (x), we have

f (-1) = (-1)⁴ + 10 (-1)³ + 35 (-1)² + 50 (-1) + 24

= 1 – 10 + 35 – 50 + 24

= 0

Therefore, (x + 1) is a factor of f (x).

Similarly, (x + 2), (x + 3) and (x + 4) are the factors of f (x).

Since, f (x) is a polynomial of degree 4. So, it cannot have more than four linear factors.

Therefore, f (x) = k (x + 1) (x + 2) (x + 3) (x + 4)

x⁴+10x³+35x²+50x +24 = k (x + 1) (x + 2) (x + 3) (x + 4)

Putting x = 0 on both sides, we get

0 + 0 + 0 + 0 + 24 = k (0 + 1) (0 + 2) (0 + 3) (0 + 4)

24 = 24k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x + 3) (x + 4), we get

f (x) = (x + 1) (x + 2) (x + 3) (x + 4)

Hence,

x⁴+10x³+35x²+50x +24 = (x + 1) (x + 2) (x + 3) (x + 4)

Let, f (x) = 2x⁴-7x³-13x²+63x-45

The factors of the constant term – 45 are

The factor of the coefficient of x⁴ is 2. Hence, possible rational roots of f (x) are:

We have,

f (1) = 2 (1)⁴ – 7 (1)³ – 13 (1)² + 63 (1) – 45

= 2 – 7 – 13 + 63 – 45

= 0

And,

f (3) = 2 (3)⁴ – 7 (3)³ – 13 (3)² + 63 (3) – 45

= 162 – 189 – 117 + 189 – 45

= 0

So, (x – 1) and (x + 3) are the factors of f (x)

(x – 1) (x + 3) is also a factor of f (x)

Let us now divide

f (x) = 2x⁴-7x³-13x²+63x-45 by (x² – 4x + 3) to get the other factors of f (x)

Using long division method, we get

2x⁴-7x³-13x²+63x-45 = (x² – 4x + 3) (2x² + x – 15)

2x⁴-7x³-13x²+63x-45 = (x – 1) (x – 3) (2x² + x – 15)

Now,

2x² + x – 15 = 2x² + 6x – 5x – 15

= 2x (x + 3) – 5 (x + 3)

= (2x – 5) (x + 3)

Hence, 2x⁴-7x³-13x²+63x-45 = (x – 1) (x – 3) (x + 3) (2x – 5)

Let, f (x) = 3x³-x²-3x+1

The factors of the constant term

The factor of the coefficient of x³ is 3. Hence, possible rational roots of f (x) are:

We have,

f (1) = 3 (1)³ – (1)² – 3 (1) + 1

= 3 – 1 – 3 + 1

= 0

So, (x – 1) is a factor of f (x)

Let us now divide

f (x) = 3x³-x²-3x+1 by (x - 1) to get the other factors of f (x)

Using long division method, we get

3x³-x²-3x+1 = (x – 1) (3x² + 2x – 1)

Now,

3x² + 2x - 1 = 3x² + 3x – x – 1

= 3x (x + 1) – 1 (x + 1)

= (3x – 1) (x + 1)

Hence, 3x³-x²-3x+1 = (x – 1) (x + 1) (3x – 1)

Let, f (x) = x³-23x²+142x-120

The factors of the constant term – 120 are

Putting x = 1, we have

f (1) = (1)³ – 23 (1)² + 142 (1) – 120

= 1 – 23 + 142 – 120

= 0

So, (x – 1) is a factor of f (x)

Let us now divide

f (x) = x³-23x²+142x-120 by (x - 1) to get the other factors of f (x)

Using long division method, we get

x³-23x²+142x-120 = (x – 1) (x² – 22x + 120)

x² – 22x + 120 = x² – 10x – 12x + 120

= x (x – 10) – 12 (x – 10)

Hence, x³-23x²+142x-120 = (x – 1) (x - 10) (x - 12)

Let, f (y) = y³-7y+ 6

The constant term in f (y) is equal to + 6 and factors of + 6 are ,

Putting y = 1 in f (y), we have

f (1) = (1)³ – 7 (1) + 6

= 1 – 7 + 6

= 0

Therefore, (y - 1) is a factor of f (y).

Similarly, (y - 2) and (y + 3) are the factors of f (y).

Since, f (y) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (y) = k (y – 1) (y - 2) (y + 3)

y³-7y+ 6 = k (y – 1) (y - 2) (y + 3)

Putting x = 0 on both sides, we get

0 – 0 + 6 = k (0 – 1) (0 - 2) (0 + 3)

6 = 6k

k = 1

Putting k = 1 in f (y) = k (y – 1) (y - 2) (y + 3), we get

f (y) = (y – 1) (y - 2) (y + 3)

Hence,

y³-7y+ 6 = (y – 1) (y - 2) (y + 3)

Let, f (x) = x³-10x²-53x-42

The factors of the constant term – 42 are

Putting x = - 1, we have

f (-1) = (-1)³ – 10 (-1)² – 53 (-1) - 42

= -1 – 10 + 53 - 42

= 0

So, (x + 1) is a factor of f (x)

Let us now divide

f (x) = x³-10x²-53x-42 by (x + 1) to get the other factors of f (x)

Using long division method, we get

x³-10x²-53x-42 = (x + 1) (x² – 11x – 42)

x² – 11x - 42 = x² – 14x + 3x - 42

= x (x – 14) + 3 (x – 14)

= (x – 14) (x + 3)

Hence, x³-10x²-53x-42 = (x + 1) (x - 14) (x + 3)

Let, f (y) = y³-2y²-29y-42

The factors of the constant term – 42 are

Putting y = - 2, we have

f (-2) = (-2)³ – 2 (-2)² – 29 (-2) - 42

= - 8 – 8 + 58 - 42

= 0

So, (y + 2) is a factor of f (y)

Let us now divide

f (y) = y³-2y²-29y-42 by (y + 2) to get the other factors of f (x)

Using long division method, we get

y³ - 2y²- 29y - 42 = (y + 2) (y² – 4y – 21)

y² – 4y - 21 = y² – 7y + 3y - 21

= y (y – 7) + 3 (y – 7)

= (y – 7) (y + 3)

Hence, y³-2y²-29y-42 = (y + 2) (y - 7) (y + 3)

Let, f (y) = 2y³-5y²-19y+42

The factors of the constant term + 42 are

Putting y = 2, we have

f (2) = 2 (2)³ – 5 (2)² – 19 (2) + 42

= 16 – 20 - 38 + 42

= 0

So, (y - 2) is a factor of f (y)

Let us now divide

f (y) = 2y³-5y²-19y+42 by (y - 2) to get the other factors of f (x)

Using long division method, we get

2y³-5y²-19y+42 = (y - 2) (2y² – y – 21)

2y² – y - 21 = (y + 3) (2y – 7)

Hence, 2y³-5y²-19y+42 = (y - 2) (2y - 7) (y + 3)

Let, f (x) = x³+13x²+32x+20

The factors of the constant term + 20 are

Putting x = -1, we have

f (-1) = (-1)³ + 13 (-1)² + 32 (-1) + 20

= -1 + 13 – 32 + 20

= 0

So, (x + 1) is a factor of f (x)

Let us now divide

f (x) = x³+13x²+32x+20 by (x + 1) to get the other factors of f (x)

Using long division method, we get

x³+13x²+32x+20 = (x + 1) (x² + 12x + 20)

x² + 2x + 20 = x² + 10x + 2x + 20

= x (x + 10) + 2 (x + 10)

= (x + 10) (x + 2)

Hence, x³+13x²+32x+20 = (x + 1) (x + 10) (x + 2)

Let, f (x) = x³-3x²-9x-5

The factors of the constant term - 5 are

Putting x = -1, we have

f (-1) = (-1)³ – 3 (-1)² – 9 (-1) - 5

= -1 – 3 + 9 - 5

= 0

So, (x + 1) is a factor of f (x)

Let us now divide

f (x) = x³-3x²-9x-5 by (x + 1) to get the other factors of f (x)

Using long division method, we get

x³-3x²-9x-5 = (x + 1) (x² - 4x 5)

x² - 4x - 5 = x² - 5x + x - 5

= x (x - 5) + 1 (x - 5)

= (x + 1) (x - 5)

Hence, x³+13x²+32x+20 = (x + 1) (x + 1) (x - 5)

= (x + 1)² (x – 5)

Let, f (y) = 2y³+y²-2y-1

The factors of the constant term - 1 are

The factor of the coefficient of y³ is 2. Hence, possible rational roots are

We have

f (1) = 2 (1)³ + (1)² – 2 (1) - 1

= 2 + 1 – 2 - 1

= 0

So, (y - 1) is a factor of f (y)

Let us now divide

f (y) = 2y³+y²-2y-1 by (y - 1) to get the other factors of f (x)

Using long division method, we get

2y³+y²-2y-1 = (y - 1) (2y² + 3y + 1)

2y² + 3y + 1 = 2y² + 2y + y + 1

= 2y (y + 1) + 1 (y + 1)

= (2y + 1) (y + 1)

Hence, 2y³+y²-2y-1 = (y - 1) (2y + 1) (y + 1)

Let, f (x) = x³-2x²-x+2

The factors of the constant term +2 are

Putting x = 1, we have

f (1) = (1)³ – 2 (1)² – (1) + 2

= 1 – 2 – 1 + 2

= 0

So, (x - 1) is a factor of f (x)

Let us now divide

f (x) = x³-2x²-x+2 by (x - 1) to get the other factors of f (x)

Using long division method, we get

x³-2x²-x+2 = (x - 1) (x² - x - 2)

x² - x - 2 = x² - 2x + x - 2

= x (x - 2) + 1 (x - 2)

= (x + 1) (x - 2)

Hence, x³-2x²-x+2 = (x - 1) (x + 1) (x - 2)

= (x - 1) (x + 1) (x – 2)

(i) Let, f (x) = x³+13x²+31x-45

Given that (x + 9) is a factor of f (x)

Let us divide f (x) by (x + 9) to get the other factors

By using long division method, we have

f (x) = x³+13x²+31x-45

= (x + 9) (x² + 4x – 5)

Now,

x² + 4x – 5 = x² + 5x – x – 5

= x (x + 5) – 1 (x + 5)

= (x – 1) (x + 5)

f (x) = (x + 9) (x + 5) (x – 1)

Therefore, x³+13x²+31x-45 = (x + 9) (x + 5) (x – 1)

(ii) Let, f (x) = 4x³+20x²+33x+18

Given that (2x + 3) is a factor of f (x)

Let us divide f (x) by (2x + 3) to get the other factors

By long division method, we have

4x³+20x²+33x+18 = (2x + 3) (2x² + 7x + 6)

2x² + 7x + 6 = 2x² + 4x + 3x + 6

= 2x (x + 2) + 3 (x + 2)

= (2x + 3) (x + 2)

4x³+20x²+33x+18 = (2x + 3) (2x + 3) (x + 2)

= (2x + 3)² (x + 2)

Hence,

4x³+20x²+33x+18 = (2x + 3)² (x + 2)

Let f(x) = x² -3ax -2a

Since, x-2 is a factor of f(x) so,

f (2) = 0

2²+ 3 a (2) - 2a =0

4 + 6a -2a = 0

a = -1

The zeros are the roots, or where the polynomial crosses the axis. A polynomial will have 2 roots that mean it has 2 zeros. To find the roots you can graph and look where it crosses the axis, or you can use the quadratic equation. This is also known as the solution.

If x =

f () = 8()³ + a ()² - 4 () + 2

0 = 1 + - 2 + 2

a = - 4

Since, x+2 is exactly divisible by f(x)

Means x+2 is a factor of f (x), so

f (-2) = 0

(-2)³ + 6 (-2)² +4 (-2) + k = 0

-16 + 24 + k = 0

k= - 8

f (x) = x³+ x² -3x +2

Given,

f (x) divided by (x+1), so reminder is equal to f (-1)

f (-1) = (-1)³ + (-1)² -3(-1) + 2

= -1 +1 +3 +2

=5

Thus, remainder is 5.

Let f (x) = x³ - 3x²a + 2a²x + b

Since, x - a is a factor of f(x)

So, f (a) = 0

a³- 3 a² (a) + 2a² (a) + b = 0

a³ - 3a³ + 2a³ + b= 0

b = 0

Let, f (x) = x³ + 4x² + 4x -3

Given f (x) is divided by x so remainder is equal to f (0)

f (0) = 0³ + 4 (0)² + 4 (0) -3

= 0 - 3

= - 3

Thus, remainder is - 3

Let f (x) = x¹⁴⁰ + 2x¹⁵¹ + k

Since, x+1 is a factor of f (x)

So, f (-1) = 0

(-1)¹⁴⁰ + 2(-1)¹⁵¹ +k = 0

1 - 2 + k=0

k = 1

Let f (x) = x³+10x² +mx + n

Since, (x + 2) and (x - 1) are factor of f (x)

So, f (-2) = 0

(-2)³ + 10 (-2)² + m (-2) + n

32 - 2m + n = 0 (i)

f (1) = 0

(1)³ + 10 (1)² + m (1) + n = 0

11 + m + n = 0 (ii)

(2) - (1)

3m -21 = 0

m = 7 (iii)

Using (iii) and (ii), we get

11 + 7 + n= 0

n = - 18

Let, f (x) = x³ + a

(x +1) is a factor of f(x), so f (-1) = 0

f (-1) =0

(-1)³ + a = 0

-1 + a = 0

a = 1

Let f(x) be a polynomial and f () = 0

x + = 2x + 1 is a factor of f (x)

f(x) = x⁴ -2x² +3x² -ax -b

Given f(x) is divided by (x-1), then remainder is 6

f (1) = 6

1⁴ - 2 (1)³ - 3 (1)² - a (1) - b = 6

1 -2 +3 -a -b = 6

-a -b = 4

a + b = - 4

Let p(x) = x³ -2(x²) + ax - b

q (x) = x² -2x -3

r (x) = x - 6

Therefore,

f(x) = p (x) – r (x)

f(x) = x³ - 2x² + ax - b - x - 6

= x³- 2x² + (a - 1) x - (b - 6)

q(x) = x² - 2x - 3

= (x + 1) (x - 3)

Thus,

(x + 1) and (x - 3) are factor of f (x)

a + b = 4

f (3) = 0

3³ - 2 (3)² + (a-1) 3 - b + 6 = 0

12 + 3a - b = 0

a = - 2, b = 6

f (x) = x⁴ +x² -20

(x² + 5) (x² - 4)

Therefore​, (x²+5) and (x²-4) are the factors of f(x)

Given,

(x-1) is a factor of f (x) but not of g(x).

Therefore, x-1 is also a factor of f(x) g(x).

Let f (x) = xⁿ +1

Since, x+ 1 is a factor of f (x), so

f (-1) = 9

Thus, n is an odd integer.

f (x) = x² + mx + 14

Since, (x +2) is a factor of f (x), so

f (-2) =0

(-2)² + m (-2) + 14 = 0

18 - 2m = 0

m = 9

Let, f(x) = x² - ax - 15

Since, (x -3) is a factor of f (x), so

f (3) = 0

3² - a (3) -15 = 0

9 - 3a - 15 = 0

a = -2

Let, p (x) = 3x³ + 8 (x)² + 8x + 3 + 5k

g (x) = x² + x + 1

Given g (x) is a factor of p (x) so remainder will be 0

Remainder= -2 + 5k

Therefore, -2 + 5k = 0

k = 2/5

We have,

(3x – 1)⁷ = a₇x⁷+a₆x⁶+a₅x⁵+….a₁x+a₀

Putting x = 1, we get

(3 * 1 – 1)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀

(2)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀

a₇+a₆+a₅+….+a₁+a₀ = 128

Let, f(x) = x⁵¹ + 51

Since, x + 1 is divided by f(x) so,

f (-1)= (-1)⁵¹ + 51

= - 1 + 51

= 50

Thus, remainder is 50

Let, f (x) = 2x² + kx

Since, x + 1 is divided by f (x) so,

f (-1)=0

2 (-1) + k (-1) = 0

k = 2

Let, f (x) = x⁴ -a²x² + 3x - 6a

Since, x + a is divided by f (x) so,

f (-a) = 0

(-a)⁴ - a² (-a)² + 3 (-a) - 6a = 0

- 9a = 0

a = 0

Since, x-1 is a factor of f (x)

Therefore,

f (1) = 0

4 (1)³ + 3 (1)² - 4 (1) + k = 0

4 + 3 - 4 + k = 0

k = - 3

Let f(x)= px² +5x +r

Since, x-2 and x-1/2 are factors of f (x)

f (2) = 0

4p + 10 + r = 0 (i)

f(1/2) = 0

p + 10 + 4r = 0 (ii)

(i) * (ii), we get,

4p + 40 + 16r = 0 (iii)

Subtracting (i) and (iii)

-30 - 15r = 0

r = - 2

Putting value of r in (i),

4p + 10 - 2 = 0

p = -2

Therefore, p = r

Let f (x) = ax⁴ + bx³ + cx² + dx + e

Since, x²- 1 is a factor of f(x)

Therefore,

f (-1) = 0

a (-1) + b (-1)³ + c (-1)² + d (-1) + e = 0

a + c + e = b + d