Polynomials

Polynomial in one variable means that the polynomial contain only one variable. For example: p(x) = x + 1, p(w) = w² + 1000, these are all polynomials in one variable.


(i)

p(x) = 4 x² - 3 x + 7

There is only one variable x with whole number power so this is a polynomial in one variable

(ii)

p(y) = y² + √2

There is only one variable y with whole number power so this is a polynomial in one variable

(iii)


There is only variable t but in 3 power of t is which is not a whole number so is not a polynomial.

(iv)


There is only one variable y but in power of y is -1 which is not a whole number so is not a polynomial

(v)

p(x, y, t) = x¹⁰ + y³ + t⁵⁰

There are three variable x, y and t and their powers are whole number so this is a polynomial in three variable and not in one variable.

Coefficient of a variable in an equation is the number that is multiplied with that variable. Therefore for finding out the coefficient of x² in the equations given below. Watch for the numbers multiplied with x² and include the positive or negative sign.

Since coefficient is the multiplicative part to a variable in a single term, then

(i) Coefficients of x² = 1

(ii) Coefficients of x² = -1

(iii) Coefficients of x² =

(iv) Coefficients of x² = 0

Monomial is a polynomial with only one term. For Example: p(x) = x

Binomial is a polynomial with two terms. for example: p(x) = 3 x + 2 x ²¹²¹²¹. So it does not matter what the power is just the number of terms should be 2.

And the degree of a polynomial is the highest power of the polynomial


Example of binomial with degree 35 is:

3x³⁵ + 7

x³⁵ + 4

Example of monomial degree is:

4x¹⁰⁰

y¹⁰⁰

(i) 5x³ has the highest power in the given polynomial which power is 3. Therefore, the degree of the polynomial is 3

(ii) –y² has the highest power in the given polynomial which power is 2. Therefore, the degree of the polynomial is 2

(iii) 5t has the highest power in the given polynomial which power is 1. Therefore, the degree of the polynomial is 1

(iv) There is no variable in the given polynomial. Therefore, the degree of the polynomial is 0

(i) Quadratic Polynomial, as its degree or highest power of variable is 2.

(ii) Cubic Polynomial, as its degree or highest power of variable is 3.

(iii) Quadratic Polynomial, as its degree or highest power of variable is 2

(iv) Linear Polynomial, as its degree or highest power of variable is 1

(v) Linear Polynomial, as its degree or highest power of variable is 1

(vi) Quadratic Polynomial, as its degree or highest power of variable is 2

(vii) Cubic Polynomial, as its degree or highest power of variable is 3.

(i)

p(x) = 5x - 4x² + 3

Now p(a) will mean replacing the variable with the value that is put in the bracket.

p (0) = 5(0) - 4(0)² + 3

p(0) = 0 - 0 + 3

p(0) = 3

(ii)

p(x) = 5x - 4x² + 3


Now p(a) will mean replacing the variable with the value that is put in bracket.

p (-1) = 5(-1) - 4(-1)² + 3

p(- 1) = - 5 - 4 + 3

p(- 1) = - 9 + 3

p(- 1) = - 6

(iii)

p(x) = 5x - 4x² + 3

Now p(a) will mean replacing the variable with the value that is put in the bracket.

p (2) = 5(2) - 4(2)² + 3

p(2) = 10 – 16 + 3

p(2) = - 6 + 3

p(2) = - 3

(i) p(y) = y² – y + 1

p (0) = (0)² – (0) + 1 = 1

p (1) = (1)² – (1) + 1 = 1

p (2) = (2)² – (2) + 1 = 3

(ii) p(t) = 2 + t + 2t² – t³

p (0) = 2 + 0 + 2 (0)² – (0)³ = 2

p (1) = 2 + (1) + 2 (1)² – (1)³

= 2 + 1 + 2 – 1 = 4

p (2) = 2 + 2 + 2 (2)² – (2)³

= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x³

p (0) = (0)³ = 0

p (1) = (1)³ = 1

p (2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)

p (0) = (0 – 1) (0 + 1) = (-1) (1) = -1

p (1) = (1 – 1) (1 + 1) = 0 (2) = 0

p (2) = (2 – 1) (2 + 1) = 1 (3) = 3

For any value of x, say a, to be a zero of polynomial p(a) = 0


Keeping that in mind,

(i) Here p(x)= 3x + 1 , for x = -1/3

Therefore, x = –1/3 is a zero of polynomial p(x) = 3x + 1

(ii) here p(x) = 5x - π, for x = 4/5,

Therefore, x = 4/5 is not a zero of polynomial p(x) = 5x - π

(iii) If x = 1 and x = -1 are zeros of polynomial p(x) = x² - 1 then p (1) and p (-1) should be 0

At, p (1) = (1)² – 1 = 0 and,

At, p (-1) = (-1)² – 1 = 0

Hence, x = 1 and -1 are zeros of polynomial p (x) = x² - 1

(iv) If x = -1 and x = 2 are zeros of polynomial p(x) = (x + 1) (x – 2) then p (-1) and p (2) should be 0

At, p (-1) = (-1 + 1) (- 1 – 2) = 0 (-3) = 0 and,

At, p (2) = (2 + 1) (2 – 2) = 3 (0) = 0

Hence, x = -1 and x = 2 are zeros of polynomial p (x) = (x + 1) (x – 2)

(v) If x = 0 is a zero of polynomial p (x) = x²

Then, p (0) should be zero

Here, p (0) = (0)² = 0

Hence, x = 0 is the zero of the polynomial p (x) = x²

(vi) If x = is a zero of the polynomial p(x) = lx + m then p () should be 0

Therefore, x = is the zero of the polynomial p(x) = l x + m

(vii) If x = and x = are zeros of the polynomial p(x) = 3 x² – 1, then

p() and p() should be 0

At, p () = 3 ()² – 1

= 3 () – 1

= 1 – 1 = 0 and,

At, p () = 3 ()² – 1

= 3 () – 1

= 4 – 1 = 3

Therefore, x = is a zero of the polynomial p(x) = 3x² + 1

But, x = is not a zero of the polynomial

(viii) If x = is a zero of polynomial p (x) = 2x + 1 then p(1/2) should be zero

At, p () = 2 () + 1

= 1 + 1 = 2

Therefore, x = is not a zero of given polynomial p(x) = 2x + 1

Zero of a polynomial means the value of the variable at which the polynomial becomes zero.

For example for a polynomial f(x), zeroes of f(x) means the values of x at which f(x) = 0

Let f(x) = x + 2, so it can clearly be seen that if you put x = - 2, then f(x) = 0.


(i)
p (x) = x + 5

p (x) = 0

x + 5 = 0

x = -5

Therefore, x = -5 is a zero of the polynomial p(x) = x + 5

(ii) p (x) = x – 5

p (x) = 0

x – 5 = 0

x = 5

Therefore, x = 5 is a zero of the polynomial p(x) = x - 5

(iii) p (x) = 2x + 5

P (x) = 0

2x + 5 = 0

2x = -5

x =

Therefore, x = is a zero of the polynomial p(x) = 2x + 5

(iv) p (x) = 3x – 2

p (x) = 0

3x – 2 = 0

x =

Therefore, x = is a zero of the polynomial p(x) = 3x - 2

(v) p (x) = 3 x

p (x) = 0

3 x = 0

x = 0

Therefore, x = 0 is a zero of the polynomial p(x) = 3x

(vi) p (x) = ax

p (x) = 0

ax = 0

x = 0

Therefore, x = 0 is a zero of the polynomial p(x) = ax

(vii) p (x) = cx + d

p (x) = 0

cx + d = 0

x =

Therefore, x = is a zero of polynomial p (x) = cx + d

By long division method, we have

Therefore, the remainder obtained is 5a when x³ – ax² + 6x – a is divided by x – a

Remainder theorem:

Now to find the remainder when x³ – ax² + 6x – a is divided by x – a

we have to put x - a = 0, Thus x = a

Let f(x) = x³ – ax² + 6x – a
and f(a) will be the remainder

f(a) = a³ - a. a² + 6 a - a
f(a) = a³ - a³ + 5 a
f(a) = 5 a


Hence 5 a is the remainder when x³ – ax² + 6x – a is divided by x – a.

To check for (3x + 7) be a factor of (3x³ + 7x)

We have to divide 3x³ + 7x by 7 + 3x

If remainder comes out be 0 then (7 + 3x) will be a factor of (3x³ + 7x)

By long division method,

As remainder is not zero so (7 + 3x) is not a factor of (3x³ + 7x)

By Remainder theorem:

put 7 + 3 x = 0

we get x = -7/3

Now for checking out the remainder put x = -7/3 in (3x³ + 7x)

we get,

f(-7/3) = 3(-7/3)³ + 7 (-7/3)

f(-7/3) = - (343/9) - (49/3)



f(-7/3) = -490/9

which is not equal to zero and hence (7 + 3x) is not a factor of (3x³ + 7x).

To check for (x + 1) as a factor of given polynomials by remainder theorem, we check for p(-1)=0.

(i) If (x + 1) is a factor of p (x) = x³ + x² + x + 1, p (-1) must be zero

Here, p (x) = x³ + x² + x + 1

p (-1) = (-1)³ + (-1)² + (-1) + 1

= -1 + 1 – 1 + 1 = 0

Therefore, x + 1 is a factor of this polynomial

(ii) If (x + 1) is a factor of p (x) = x⁴ + x³ + x² + x + 1, p (-1) must be zero

Here, p (x) = x⁴+ x³ + x² + x + 1

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

= 1 - 1 + 1 – 1 + 1 ≠ 0

Therefore, x + 1 is not a factor of this polynomial

(iii) If (x + 1) is a factor of p (x) = x⁴ + 3x³ + 3x² + x + 1, p (-1) must be zero

Here, p (x) = x⁴+ 3x³ + 3x² + x + 1

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

= 1 - 3 + 3 – 1 + 1 ≠ 0

Therefore, x + 1 is not a factor of this polynomial

(iv) If (x + 1) is a factor of polynomial

p (x) = x³ – x² – (2 + √2)x + √2, p(-1) must be zero

p (-1) = (-1)³ – (-1)² – (2 + √2) (-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2

As, p (-1) ≠ 0

Therefore, x + 1 is not a factor of this polynomial

(i) If g (x) = x + 1 is a factor of given polynomial p (x), p (-1) must be zero

p (x) = 2x³ + x² – 2x – 1

p (-1) = 2 (-1)³ + (-1)² – 2 (-1) – 1

= 2 (-1) + 1 + 2 – 1 = 0

Hence, g (x) = x + 1 is a factor of given polynomial

(ii) If g (x) = x + 2 is a factor of given polynomial p (x), p (-2) must be zero

p (x) = x³ + 3x² + 3x + 1

p (-2) = (-2)³ + 3(-2)² + 3 (-2) + 1

= -8 + 12 - 6 + 1 = -1

As, p (-2) ≠ 0

Hence, g (x) = x + 2 is not a factor of given polynomial

(iii) If g (x) = x – 3 is a factor of given polynomial p (x), p (3) must be zero

p (x) = x³ – 4x² + x + 6

p (3) = (3)³ – 4 (3)² + 3 + 6

= 27 – 36 + 9 = 0

Therefore, g(x) = x – 3 is a factor of the given polynomial

If g(x) is a factor of another polynomial p(x) then, the points at which g(x) = 0, p(x) will also be equal to 0

for example: Let p(x) = x² - 1 and g(x) = x + 1

By remainder theorem.

Now if x + 1 is a factor of p(x), then x + 1 = 0, x = -1 at this value p(x) = 0

Thus, p(- 1) = 0


(i) If x – 1 is a factor of polynomial p (x) = x² + x + k, then

x - 1 = 0

x = 1

Therefore,

p (1) = 0

(1)² + 1+ k = 0

2 + k = 0

k = -2

Therefore, value of k is -2

(ii) If x – 1 is a factor of polynomial p (x) = 2x² + kx + , then

p (1) = 0

2(1)² + k (1) + = 0

2 + k + = 0

k = -2 -

= - (2 + )

Therefore, value of k is – (2 + )

(iii) If x – 1 is a factor of given polynomial p(x) = kx² - x + 1, then

p (1) = 0

k (1)² - (1) + 1 = 0

k - + 1 = 0

k = - 1

Therefore, value of k is √2 – 1

(iv) If x – 1 is a factor of the given polynomial p(x) = kx² – 3x + k, then

p (1) = 0

k(1)² + 3(1) + k = 0

k – 3 + k = 0

2k – 3 = 0

k =

Therefore, value of k is

For quadratic polynomial. We factorize the middle term such that the product of the terms is equal to the product of the last term and coefficient of x² and sum or difference is equal to the middle term.

(i)

12x² - 7x + 1

For the equation given above:

7 is to be factorized such that, the product of two terms is 12 x 1 = 12 and the sum is equal to 7
This can be done by 3 and 4
as 3 x 4 = 12
and 3 + 4 = 7
Therefore,

12x² - 7x + 1 = 12x² – 4x – 3x + 1

= 4x (3x – 1) – 1 (3x – 1)

= (3x – 1) (4x – 1)

(ii)

2x² + 7x + 3

For the equation given above:

7 is to be factorized such that, the product of two terms is 3 x 2 = 6 and the sum is equal to 7
This can be done by 6 and 1
as 6 x 1 = 6
and 6 + 1 = 7
Therefore,

2x² + 7x + 3 = 2x² + 6x + x + 3

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

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

(iii)

6x² + 5x – 6

6x² + 5x – 6 = 6x² + 9x – 4x – 6

= 3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

(iv)

3x² - x - 4

For the equation given above:

7 is to be factorized such that, the product of two terms is 4 x 3 = 12 and the difference = 1
This can be done by 4 and 3
as 4 x 3 = 12
and 4 - 3 = 1
Therefore,

3x² - x - 4=3x² – 4x + 3x – 4

= x (3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

(i) Let p(x) = x³ – 2x² – x + 2

By trial method, we find that

p (-1) = (-1)³ – 2 (-1)² – (-1) + 2

= -1 – 2 + 1 + 2 = 0

Therefore, x= -1 satisfies the equation and (x + 1) is the factor of p (x)
Now, we divide p(x) = x³ - 2x - x + 2 by (x + 1)

Now, Dividend = (Divisor × Quotient) + Remainder

p (x) = (x + 1) (x² – 3x + 2)

= (x + 1) (x² – x – 2x + 2)

= (x + 1) {x (x – 1) – 2 (x – 1)}

= (x + 1) (x – 1) (x - 2)

(ii) Let p (x) = x³ – 3x² – 9x – 5

By trial method,

p (5) = (5)³ – 3 (5)² – 9 (5) – 5

= 125 – 75 – 45 – 5 = 0

Therefore, (x – 5) is the factor of p (x)

Now, Dividend = (Divisor × Quotient) + Remainder

p(x) = (x – 5) (x² + 2x + 1)

= (x – 5) (x² + x + x + 1)

= (x – 5) {x (x + 1) + 1 (x + 1)}

= (x – 5) (x + 1) (x + 1)

(iii) Let p (x) = x³ + 13x² + 32x + 20

By trial method,

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

= -1 + 13 – 32 + 20 = 0

Therefore, x = -1 satisfies the equation p(x) = 0 so, (x + 1) is the factor of p (x)

Now, Dividend = (Divisor × Quotient) + Remainder

p(x) = (x + 1) (x² + 12x + 20)

= (x + 1) (x² + 2x + 10x + 20)

= (x + 1) {x (x + 2) + 10 (x + 2)}

= (x + 1) (x + 2) (x + 10)

(iv) Let, p(y) = 2y³ + y² – 2y – 1

By trial method,

p (1) = 2 (1)³ + (1)² – 2 (1) – 1

= 2 + 1 -1 -1 = 0

Therefore, y = 1 satisfies the equation p(y) = 0 so, (y – 1) is a factor of p(y)
Now, we divide p(y) = 2y³ + y² – 2y – 1 by (y - 1)

Now, Dividend = (Divisor × Quotient) + Remainder

p(y) = (y – 1) (2y² - 3y + 1)

= (y – 1) (2y² - 2y - y + 1)

= (y – 1) {2y (y - 1) - 1 (y - 1)}

= (y – 1) (2y - 1) (y - 1)

(i) (x + 4) (x + 10)

Using identity,

(x + a) (x + b) = x² + (a + b)x + ab

In (x + 4) (x + 10), a = 4 and b = 10

Now,

(x + 4) (x + 10) = x² + (4 + 10)x + (4 × 10)

= x² + 14x + 40

(ii) (x + 8) (x - 10)

Using identity,

(x + a) (x + b) = x² + (a + b) x + ab

Here, a = 8 and b = -10

(x + 8) (x – 10) = x² + {8 + (-10)} x + {8 × (-10)}

= x² + (8 – 10)x – 80

= x² – 2x – 80

(iii) (3x + 4) (3x - 5)

Using identity,

(x + a) (x + b) = x² + (a + b) x + ab

Here, x = 3x, a = 4 and b = -5

(3x + 4) (3x – 5) = (3x)² + {4 + (-5)} 3x + {4 × (-5)}

= 9x² + 3x (4 – 5) – 20

= 9x² – 3x – 20

(iv)

Using identity,

(x + y) (x – y) = x² – y²

Here, x = y² and y =

(v) (3 - 2x) (3 + 2x)

Using identity,

(x + y) (x – y) = x² – y²

Here, x = 3 and y = 2x

(3 – 2x) (3 + 2x) = 3² – (2x)²

= 9 – 4x²

(i) Using identity,

(x + a) (x + b) = x² + (a + b) x + ab

Here, x = 100, a = 3 and b = 7

103 × 107 = (100 + 3) (100 + 7)

= (100)² + (3 + 7) 100 + (3 × 7)

= 10000 + 1000 + 21

= 11021

(ii) Using identity,

(x - a) (x - b) = x² - (a + b) x + ab

Here, x = 100, a = 5 and b = 4

95 × 96 = (100 - 5) (100 - 4)

= (100)² - (5 + 4) 90 + (5 x 4)

= 10000 - 1800 + 20

= 9120

(iii) Using identity,

(x + y) (x – y) = x² – y²

Here, x = 100 and y = 4

104 × 96 = (100 + 4) (100 – 4)

= (100)² – (4)²

= 10000 – 16

= 9984

(i) 9x² + 6xy + y² = (3x)² + (2 × 3x × y) + y²

Using identity, (a + b)² = a² + 2ab + b²

Here, a = 3x and b = y

9x² + 6xy + y² = (3x)² + (2 × 3 × y) + y²

= (3x + y)²

= (3x + y) (3x + y)

(ii) 4y² - 4y + 1 = (4y)² - (2 × 2y × 1) + 1²

Using identity, (a - b)² = a² - 2ab + b²

Here, a = 2y and b = 1

4y² - 4y + 1 = (2y)² - (2 × 2y × 1) + 1²

= (2y - 1)²

= (2y - 1) (2y - 1)

(iii) Using identity,

a² – b² = (a + b) (a – b)

Here, a = x and b = ()

(i) Using identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = x, b = 2y and c = 4z

(x + 2y + 4z)² = x² + (2y)² + (4z)² + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)

= x² + 4y² + 16z² + 4xy + 16yz + 8xz

(ii) Using identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = 2x, b = -y and c = z

(2x – y + z)² = (2x)² + (-y)² + z² + (2 × 2x × -y) + (2 × -y × z) + (2 × z × 2x)

= 4x² + y² + z² – 4xy – 2yz + 4xz

(iii) Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = -2x, b = 3y and c = 2z

(-2x + 3y + 2z)² = (-2x)² + (3y)² + (2z)² + (2 × -2x × 3y) + (2 × 3y × 2z) + (2 × 2z × -2x)

= 4x² + 9y² + 4z² – 12xy + 12yz – 8xz

(iv) Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = 3a, b = -7b and c = -c

(3a – 7b – c)² = (3a)² + (-7b)² + (-c)²+ (2 ×3a × -7b) + (2 × -7b × -c) + (2 × -c × 3a)

= 9a² + 49b² + c² – 42ab + 14bc – 6ac

(v) Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = -2x, b = 5y and c = -3z

(-2x + 5y – 3z)² = (-2x)² + (5y)² + (-3z)²+ (2 ×-2x × 5y) + (2 × 5y × -3z) + (2 × -3z × -2x)

= 4x² + 25y² + 9z² – 20xy -30yz + 12xz

(vi) Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, a = a, b = b and c = 1

(a – b + 1)² = (a)² + (-b)² + (1)²+ (2 ×a × -b) + (2 × -b × 1) + (2 × 1 × a)

= a² + b² + 1 – ab - b + a

(i) Using identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

4x² + 9y² + 16z² + 12xy – 24yz – 16xz

= (2x)² + (3y)² + (-4z)² + (2 × 2 × 3y) + (2 × 3y × -4z) + (2 × -4z × 2x)

= (2x + 3y – 4z)²

= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) Using identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

2x² + y² + 8z² - 2√2xy + 4√2yz – 8xz

= (-√2x)² + (y)² + (2√2z)² + (2 × -√2x × y) + (2 × y × 2√2z) + (2 × 2√2z × -√2x)

= (-√2x + y + 2√2z)²

= (-√2x + y + 2√2z) (-√2x + y + 2√2z)

(i) (2x + 1)³

Using identity,

(a + b)³ = a³ + b³ + 3ab (a + b)

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

= 8x³ + 1 + 6x (2x + 1)

= 8x³ + 12x² + 6x + 1

(ii) Using identity,

(a – b)³ = a³ – b³ – 3ab (a – b)

(2a – 3b)³ = (2a)³ – (3b)³ – (3 × 2a × 3b) (2a – 3b)

= 8a³ – 27b³ – 18ab (2a – 3b)

= 8a³ – 27b³ – 36a²b + 54ab²

(iii) Using identity,

(a + b)³ = a³ + b³ + 3ab (a + b)

(x + 1)³ = (x)³ + 1³ + (3 × x × 1) (x + 1)

= x³ + 1 + x (x + 1)

= x³ + 1 + x² + x

= x³ + x² + x + 1

(iv) Using identity,

(a – b)³ = a³ – b³ – 3ab (a – b)

(x - y)³ = (x)³ – (y)³ – (3 × x × y) (x - y)

= x³ - y³ – 2xy (x - y)

= x³ - y³ – 2x²y + xy²

(i) (99)³ = (100 – 1)³

Using identity,

(a – b)³ = a³ – b³ – 3ab(a – b)

(100 – 1)³ = (100)³ – 1³ – (3 × 100 × 1) (100 – 1)

= 1000000 – 1 – 300 (100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)³ = (100 + 2)³

Using identity,

(a + b)³ = a³ + b³ + 3ab (a + b)

(100 + 2)³ = (100)³ + 2³ + (3 × 100 × 2) (100 + 2)

= 1000000 + 8 + 600 (100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)³ = (1000 – 2)³

Using identity,

(a – b)³ = a³ – b³ – 3ab (a – b)

(1000 – 2)³ = (1000)³ – 2³ – (3 × 1000 × 2) (1000 – 2)

= 1000000000 – 8 – 6000 (1000 – 2)

= 1000000000 – 8 – 6000000 + 12000

= 994011992

(i)
Using identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

8a³ + b³ + 12a²b + 6ab²

= (2a)³ + b³ + 3 (2a) (2b) + 3 (2a) (b)²

= (2a + b)³

= (2a + b) (2a + b) (2a + b)

(ii)
Using identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

8a³ – b³ – 12a²b + 6ab²

= (2a)³ – b³ – 3 (2a)²b + 3 (2a) (b)²

= (2a – b)³

= (2a – b) (2a – b) (2a – b)

(iii)
Using identity,

(a - b)³ = a³ - b³ - 3a²b + 3ab²

27 – 125a³ – 135a + 225a²

= 3³ – (5a)³ – 3 (3)²(5a) + 3 (3) (5a)²

= (3 – 5a)³

= (3 – 5a) (3 – 5a) (3 – 5a)

(iv)
Using identity,

(a - b)³ = a³ - b³ - 3a²b + 3ab²

64a³ – 27b³ – 144a²b + 108ab²

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

= (4a – 3b)³

= (4a – 3b) (4a – 3b) (4a – 3b)

(v)
Using identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

27p³ - - p² + p

= (3p)³ – ()³ – 3 (3p)² () + 3 (3p) ()²

= (3p - )³

= (3p - ) (3p - ) (3p - )

(i)x³ + y³ = (x + y) (x² + y² – xy)

We know that,

(x + y)³ = x³ + y³ + 3xy (x + y)

⇒ x³ + y³ = (x + y)³ – 3xy (x + y)

[Taking (x + y) common]


⇒ x³ + y³ = (x + y) [(x + y)² – 3xy)

⇒ x³ + y³ = (x + y) [(x² + y² + 2xy) – 3xy] ........[(x + y)² = x² + y² + 2 x y]

⇒ x³ + y³ = (x + y) (x² + y² – xy)

(ii)x³ - y³ = (x - y) (x² + y² + xy)

We know that,

(x - y)³ = x³ - y³ - 3xy (x - y)

⇒ x³ - y³ = (x - y)³ + 3xy (x - y)

⇒ x³ - y³ = (x - y) [(x - y)² + 3xy)

⇒ x³ - y³ = (x - y) [(x² + y² - 2xy) + 3xy] ......[(x - y)² = x² + y² - 2 x y]

= x³ - y³ = (x - y) (x² + y² + xy)

(i) Using identity,

x³ + y³ = (x + y) (x² – xy + y²)

27y³ + 125z³ = (3y)³ + (5z)³

= (3y + 5z) {(3y)² – (3y) (5z) + (5z)²}

= (3y + 5z) (9y² – 15yz + 25z²)

(ii) Using identity,

x³ – y³ = (x – y) (x² + xy + y²)

64m³ – 343n³ = (4m)³ – (7n)³

= (4m - 7n) {(4m)² + (4m) (7n) + (7n)²}

= (4m - 7n) (16m² + 28mn + 49n)²

27x³ + y³ + z³ – 9xyz = (3x)³ + y³ + z³ – 3 × 3xyz

Using identity,

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

27x³ + y³ + z³ – 9xyz

= (3x + y + z) {(3x)² + y² + z² – 3xy – yz – 3xz}

= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

We know that,

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

Dividing and Multiplying RHS by 2 gives,

As, [ (x - y)² = x² + y² - 2x y, (y - z)² = y² + z² - 2y z, (z - x)² = z² + x² - 2z x]

= R.H.S

Hence, proved.

We know that,

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

Now, put (x + y + z) = 0

x³ + y³ + z³ – 3xyz = (0) (x² + y² + z² – xy – yz – xz)

x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz
Hence proved

(i) Let x = -12, y = 7 and z = 5

We observed that,

x + y + z = -12 + 7 + 5 = 0

We know that if,

x + y + z = 0

Then,

x³ + y³ + z³ = 3xyz

(-12)³ + (7)³ + (5)³ = 3 (-12) (7) (5)

= -1260

(ii) Let x = 28, y = -15 and z = -13

We observed that,

x + y + z = 28 – 15 – 13 = 0

We know that if,

x + y + z = 0

Then,

x³ + y³ + z³ = 3xyz

(28)³ + (-15)³ + (-13)³ = 3 (28) (-15) (-13)

= 16380

(i) Area: 25a² – 35a + 12

Since the area is the product of length and breadth therefore by factorizing the given area, we can know the length and breadth of the rectangle

25a² – 35a + 12

= 25a² – 15a – 20a + 12

= 5a (5a – 3) – 4 (5a – 3)

= (5a – 4) (5a – 3)

Possible expression for length = (5a – 4) or (5 a - 3)

Possible expression for breadth = (5a – 3) or ( 5 a - 4)

(ii) Area: 35y² + 13y – 12

35y² + 13y – 12

= 35y² – 15y + 28y – 12

= 5y (7y – 3) + 4 (7y – 3)

= (5y + 4) (7y – 3)

Possible expression for length = (5y + 4) or (7 y - 3)

Possible expression for breadth = (7y – 3) or (5 y + 4)

(i) Volume: 3x² – 12x

Since, the volume is the product of length, breadth, and height therefore by factorizing the given volume we can know the length, breadth and height of the cuboid

3x² – 12x

= 3x (x – 4)

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x – 4)

(ii) Volume: 12ky² + 8ky – 20k

Since, the volume is the product of length, breadth, and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid

12ky² + 8ky – 20k

= 4k (3y² + 2y – 5)

= 4k (3y² + 5y – 3y – 5)

= 4k [y (3y + 5) – 1 (3y + 5)]

= 4k (3y + 5) (y – 1)

Possible expression for length = 4k

Possible expression for breadth = (3y + 5)

Possible expression for height = (y – 1)