Polynomials

(i)
factorize the given polynomial by splitting the middle term:

⇒ x² - 4x + 2x – 8

⇒ x (x - 4) + 2 (x - 4)

For zeros of f(x),
f(x) = 0

⇒(x + 2) (x - 4) = 0
x+2=0
x=-2
x-4=0
x=4

⇒x = -2, 4

Therefore zeros of the polynomial are -2 & 4
In a polynomial the relations hold are as follows:
sum of zeroes is equal to
product of zeroes is equal to
For the given polynomial,

Sum of zeros = -2 + 4 = 2
And is -(-2) = 2
Hence the value of and sum of zeroes are same.

Product of zeros = -2 × 4 = -8
is -8.
Hence the value of and product of zeroes are same.

(ii)
factorize the given polynomial by splitting the middle term:

⇒ 4s² -2s - 2s + 1

⇒ 2s (2s - 1) -1 (2s - 1)

For zeros of g(s), g(s) = 0

(2s - 1) (2s - 1) = 0
2s - 1=0

s =

Therefore zeros of the polynomial are ,
In a polynomial the relations hold are as follows:
sum of zeroes is equal to
product of zeroes is equal to
For the given polynomial,

Sum of zeros = + = 1

Hence the value of and sum of zeroes are same.
Product of zeros = × =

Hence the value of and product of zeroes are same.

(iii)
use the formula to solve the above equation,
^{Here a is t and b is .
Solve the given expression as:}

For zeros of h(t),
h(t) = 0

Therefore zeros of the given polynomial are t = √15 & -√15
In a polynomial the relations hold are as follows:
sum of zeroes is equal to
product of zeroes is equal to
For the given polynomial,

Sum of zeros = √15 + (- √15) = 0
The value of is 0.
Hence, the value of and sum of zeroes are same.

Product of zeros
The value of is -^{.
Hence the value of and product of zeroes are same.}

(iv) f(x) =
Write the equation in the form of ax² +bx+c as:

6x² - 7x -3
factorize the given polynomial by splitting the middle term:

⇒ 6x² - 9x + 2x - 3

⇒ 3x(2x - 3) +1(2x - 3)

⇒ (3x + 1) (2x - 3)

For zeros of f(x),
f(x) = 0

⇒ (3x + 1) (2x - 3) = 0

x =

Therefore zeros of the polynomial are
In a polynomial the relations hold are as follows:
sum of zeroes is equal to
product of zeroes is equal to

Sum of zeros = = = = =

Product of zeros = × = = =

(v)

P (x) = x² + 3√2x - √2x - 6

For zeros of p(x), p(x) = 0

⇒ x (x + 3√2) -√2 (x + 3√2) = 0

⇒ (x - √2) (x + 3√2) = 0

x = √2, -3√2

Therefore zeros of the polynomial are √2 & -3√2

Sum of zeros = √2 -3√2 = -2√2 = -2√2 =

Product of zeros = √2 × -3√2 = -6 = -6 =

(vi) q (x) = √3x² + 10x + 7√3

⇒ √3x² + 10x + 7√3

⇒ √3x² + 7x + 3x + 7√3

⇒ √3x (x +) + 3 (x + )

⇒ (√3x + 3) (x +)

For zeros of Q(x), Q(x) = 0

(√3x + 3) (x +) = 0

X = ,

Therefore zeros of the polynomial are ,

Sum of zeros = + =

Product of zeros = = × = 7 =

(vii) f(x) = x² - (√3 + 1)x + √3

f(x) = x² - √3x - x + √3

f(x) = x(x - √3) -1(x - √3)

f(x) = (x - 1) (x - √3)

For zeros of f(x), f(x) = 0

(x - 1) (x - √3) = 0

X = 1, √3

Therefore zeros of the polynomial are 1 & √3

Sum of zeros = 1 + √3 = √3 + 1=

Product of zeros = 1 × √3 = √3=

(viii) g(x) = a(x² + 13) – x(a² + 1)

g(x) = ax² - a²x – x + a
g(x) = ax² - (a² + 1)x + a

g(x) = ax(x - a) -1(x - a)

g(x) = (ax - 1) (x - a)

For zeros of g(x), g(x) = 0

(ax - 1) (x - a) = 0

X = , a

Therefore zeros of the polynomial are & a

Sum of zeros

Product of zeros = × a = 1 = 1 =

α and β are the zeros of the quadratic polynomial

Sum of the roots = = = =

Product of the roots = = =

= ()

On substituting values from above, we get

= ×
=

Let and β are the roots of the given eqn

Sum of the roots = = = = 4

Product of the roots = = = = 3

Now, to evaluate

α⁴β³ + α³β⁴ = α³β³(α + β)

On substituting values from above, we get

⇒ 3³× 4 = 108

Given: α and β are the zeros of the quadratic polynomial

To find: the value of .

Solution:

and β are the roots of the given eqn.

We know,

Sum of the roots =


⇒ = = 5


And
Product of the roots = = = = 4

⇒ = = 4



Now,

,

On substituting values from above, we get

=

Let and β are the roots of the given eqn

Sum of the roots = = = =

Product of the roots = = =

On substituting values from above, we get

= = = = 7

Let and β are the roots of the given eqn

Sum of the roots = = = = 1

Product of the roots = = = = -4

On substituting values from above, we get

Let and β are the roots of the given eqn

Sum of the roots = = =

Product of the roots = = = =

= = Using (a + b)² = a² + b² + 2ab

On substituting values from above, we get

⇒ =

Let and β are the roots of the given eqn

Sum of the roots = = = -(-6/3) = 2

Product of the roots = = =

= +3 β = +3 β

Using (a + b)² = a² + b² + 2ab

Given : α and β are the zeros of the quadratic polynomial
To find : the value of .
Solution : and β are the roots of the given eq.
Sum of root:

Product of the roots:


Now,

Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is

Sum of the roots = = = = 2k

= 2k

0 = 2k

k = 0

Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

Sum of the roots = = =

Product of the roots = = = = 3

According to the question: sum of zeros = product of zeros

= 3

k =

Consider ,

Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

Sum of the roots = = = = -p

Product of the roots = = = 45

According to the question: squared difference of the zeros = 144

= 144

⇒ (-p)² - 4× 45 = 144

⇒ p² = 324

⇒ p = ±18

Given:α and β are the zeros of the quadratic polynomial

To show: ..... (1)

solution:

one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

Sum of the roots is:





Product of coefficient is:

⇒ = +1

On substituting values, we get


= -(p+c) + p + 1

⇒ =

⇒ =

Hence proved

A quadratic equation when sum and product of its zeros is given by:

f(x) = k{x² - (sum of zeros)x + product of the zeros}, where k is a constant

α + β = 24 ....(1)

α - β = 8 ....(2)

Adding 1 and 2 we get,

α + β + α - β = 24 + 8

⇒ 2α = 32

⇒ α = 16

Substitute value in 1 to get

16 + β = 24

⇒ β = 24-16

⇒ β = 8

α = 16 and β = 8

f(x) = k{x² - (24)x + 16 × 8}

f(x) = k(x² - 24x + 128)

A quadratic equation when sum and product of its zeros is given by:

, where k is a constant

Consider the polynomial ,

Sum of the roots = = = =

Product of the roots = = =

⇒ Sum of the zeros of new eq =



=

=




[Using (a + b)² = a² + b² + 2ab]

⇒

=

= -4

Product of the zeros of new eqn = = 4

)

A quadratic equation when sum and product of its zeros is given by:

, where k is a constant

Sum of the roots = = = =

Product of the roots = = =

Sum of the zeros of new eqn = =

⇒






Now,



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

⇒ = =

Product of the zeros of new eqn = =

⇒












= =

)

A quadratic equation when sum and product of its zeros is given by:

, where k is a constant

Sum of the roots = = = =

Product of the roots = = =

Sum of the zeros of new eqn = = =

⇒ 2×-2× = 2-) [Using (a + b)² = a² + b² + 2ab]

Product of the zeros of new eqn = =

⇒

(i)

A quadratic equation when sum and product of its zeros is given by:

, where k is a constant

Sum of the roots = = = =

Product of the roots = = =

Sum of the zeros of new eqn = =

Product of the zeros of new eqn = =

(ii) A quadratic equation when sum and product of its zeros is given by:

, where k is a constant

Sum of the roots = = = =

Product of the roots = = =

Sum of the zeros of new eqn = = = = =

Product of the zeros of new eqn = = = = =

Therefore eqn is:

= = = = 0

(i) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

Sum of the roots = = =

Product of the roots = = =

⇒ On substituting values, we get

⇒ = =

(ii) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

⇒ On substituting values, we get

⇒ =

=

(iii) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

⇒ On substituting values, we get

⇒

(iv) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

⇒ On substituting values, we get

⇒=

(v) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

⇒= [Using (a + b)² = a² + b² + 2ab]

On substituting values, we get

⇒=

⇒=

⇒=

(vi) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

= =

⇒ On substituting values, we get

=

⇒ =

(vii) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

=

⇒ = =

⇒ =

(viii) Let one root of the given quadratic polynomial is

Other root of the given quadratic polynomial is β

f(x) = ax² + bx + c

Sum of the roots = = =

Product of the roots = = =

=

⇒ + = =

On substituting values, we get

=

= = b

_(i)



=

= = 0

= = 0

Let = ; β=1; γ=-2

β + γ = = =

β+ βγ + γ = = =

βγ = = -1 =

_(ii)


=

= = 0

= = 0

Let =2; β=1; γ=1

β + γ = = 4 =

β+ βγ + γ = =

βγ = = 2 =

A quadratic equation when sum and product of its zeros is given by:

where k is any non-zero real number.

⇒

⇒ f, where k is any non-zero real number.

Let are the zeros of the given polynomial.

Sum of the zeros =

⇒ =

⇒ = =

⇒ =

= = = 15

= 15

= 15

On substituting =

= 15

= =

= ±

=

=

Let are the zeros of the given polynomial.

Sum of the zeros =

⇒ =

⇒ =

⇒ =

Since is the zero of the polynomial, therefore

⇒ a³ + 3pa² + 3qa + r = 0

=, we get

⇒ -p³ + 3p³ - 3pq + r = 0
⇒ 2p³ - 3pq + r = 0

Let are the zeros of the given polynomial.

Sum of the zeros =

⇒ =

⇒ =

⇒ =

Since is the zero of the polynomial, therefore

=, we get

⇒

⇒

Let are the zeros of the given polynomial.

Sum of the zeros =

⇒ =

⇒ =

⇒ =

Since is the zero of the polynomial, therefore

On substituting =

⇒

(i) and

Degree of ; therefore degree of and degree of remainder is less than 2,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , ,

On substituting these values for

(ii) and

Degree of ; therefore degree of and degree of remainder is less than 2.

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , , ;

On substituting these values for

(iii) and

Degree of ; therefore degree of and degree of remainder is less than 2,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , ,

On substituting these values for

(iv) and

Degree of ; therefore degree of and degree of remainder is less than 2,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , ,

On substituting these values for

(i) and

Degree of ; therefore degree of and degree of remainder is of degree 1 or less,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , , ,

On substituting these values for

Since remainder is zero, therefore

(ii) and

Degree of ; therefore degree of and degree of remainder is of degree 1 or less,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , , ,

On substituting these values for

Since remainder is 2, therefore

(iii) and

Degree of ; therefore degree of and degree of remainder is of degree 2 or less,

Let and

By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

On substituting values in the above relation we get,

On comparing coefficients we get,

On solving above equations we get,

, , , ,

On substituting these values for

Since remainder is , therefore

We know that if is a zero of a polynomial then

Since -2 and -1 are zeros of Therefore is a factor of .

Now on dividing to find other zeros.

By applying division algorithm, we have:

= ()()

= ()

= {}

Hence, the zeros of the given polynomial are:

We know that if is a zero of a polynomial then

Since -2 is zero of Therefore is a factor of .

Now on divide by to find other zeros.

By applying division algorithm, we have:

x³ + 13x² + 32x + 20 = (x+2)(x²+11x+10)

We do factorisation here by splitting the middle term,

⇒ x³ + 13x² + 32x + 20 = (x+2) {x(x+10)+1(x+10)}

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

Hence, the zeros of the given polynomial are:

We know that if is a zero of a polynomial then

Since -√3 and √3 are zeros of Therefore and are factors of .

Now on dividing to find other zeros.

By applying division algorithm, we have:

= ()

= ()

= {}

Hence, the zeros of the given polynomial are:

We know that if is a zero of a polynomial then

Since and are zeros of Therefore and are factors of .

Now on dividing ⇒ to find other zeros.

By applying division algorithm, we have:

= ()

= ()

= {}

Hence, the zeros of the given polynomial are:

_{By division Algorithm we have,}

From above relation we got that if we add then resulting polynomial is divisible by .

Now find the remainder when is divisible by .

Therefore has to be added so that resulting polynomial is divisible by

Given : the polynomial
To find : What must be subtracted from the polynomial so that the resulting polynomial is exactly divisible by
Solution :
Let g(x) be
By applying division algorithm:

Dividend = Quotient× Divisor + Remainder

Dividend - Remainder = Quotient× Divisor

Now find the remainder when is divisible by .

Therefore has to be subtracted so that resulting polynomial is divisible by

We know that if is a zero of a polynomial then

Since -2 and 2 are zeros of Therefore and are factors of .

Now on dividing f(x) by (x - 2)(x + 2) = x² - 4 to find other zeros.

By applying division algorithm, we have:

Hence, two other zeroes are -6 and 5

We know that if is a zero of a polynomial then

Since -√2 and √2 are zeros of Therefore and are factors of .

Now on dividing to find other zeros.

By applying division algorithm, we have:

= (2)

= ()

= {}

Hence, the zeros of the given polynomial are:

We know that if is a zero of a polynomial then

Since -√3 and √3 are zeros of Therefore and are factors of .

Now on dividing to find other zeros.

By applying division algorithm, we have:

= ()

Hence, the zeros of the given polynomial are:

We know that if is a zero of a polynomial then

Since -√2 and √2 are zeros of Therefore and are factors of .

Now on dividing to find other zeros.

By applying division algorithm, we have:

= ()

Hence, the zeros of the given polynomial are:

Given,

f(x) = x² + x + 1

α = a

β = b

c = c

α + β = – b/a

= – 1/1

= – 1 …….. Equ.(i)

αβ = c/a = 1 ….. Equ.(ii)

Dividing (i) from (ii)

We get,

– 1/1 = – 1

So,

1/α + 1/β = – 1

A polynomial is a mathematical expression containing a sum of powers in one or more variables multiplied by coefficients or we can say an expression of more than two algebraic terms that contain different powers of the same variable.
But;
• Not divisible by a variable.
• A variable's exponents can only be 0,1,2,3,... etc.
• It can't have an infinite number of terms.

Example - 5xy² – 3x + 5y³ – 3

And a polynomial with real coefficients is a product of irreducible polynomials of first and second degrees or in simple words, a polynomial having only real numbers as coefficients is the real coefficient Polynomial.

A degree in a polynomial function is the greatest exponent of that equation, which determines the most number of solutions that a function could have. Or in simple words the degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.

Each equation contains from one to several terms, which are divided by numbers or variables with different exponents.

For Example,

y = 3x¹³ + 5x³

As we can see it has two terms,

3x¹³ and 5x³

And the degree of the polynomial is 13, and that's the highest degree of any term in the equation.

Given,

p(x) = 4x² + 3x + 7

α + β = – b/a

= – 3x/4x² = – 3/4x ………. (i)

αβ = c/a

= 7/4x² …………. (ii)

Dividing (i) from (ii),

We get,

So,

1/α + 1/β = – 3/7

Given;

f(x) = (k² + 4) x² + 13x + 4k,

One zero of the polynomial is reciprocal of the other,

Let a be the one zero,

∴ The other zero will be 1/a

As we know that,

Product of the zeros = c/a = 4k/k² + 4

∴ 4k/k² + 4 = 1

⇒ 4k = k² + 4

⇒ k² + 4 – 4k = 0

⇒ (k – 2)² = 0

⇒ k = 2

So the value of k is 2

As we know the condition for linear polynomial degree equals to 1

So,

f(x) = ax¹ + b,

Where,

a 0

Given,

f(x) = 2x³ – 3kx² + 4x – 5

Sum of the zeros of the polynomial = 6

Let x, y and z be the zeroes than,

x + y + z = 6……Equation (i)

So,

x + y + z = – b/a

= – ( – 3k)/2

From Eq. (i) we get,

3k/2 = 6

k/2 = 6/3

k = 2×2 = 4

To write the standard form of a quadratic polynomial with real coefficients;

Let’s take a polynomial,

f (x) = ax² + bx + c,

As we know that the quadratic polynomial = 2

So,

a 0

To write the standard form of a cubic polynomial with real coefficients

Let’s take,

f(x) = ax³ + bx² + cx + d,

Where a 0

Because degree of a cubic polynomial is 3

Given: If α and β are the zeros of the polynomial f(x) = x² + px + q,

To find: a polynomial having and is its zeros is


Solution:

f(x) = x² + px + q

⇒ α + β = – p

and αβ = q


Let S and P denote respectively the sum and product of zeroes of the required polynomial,

So,



...... (1)

And

...... (2)



Put the values of α + β and αβ in (1) and (2) to get,



And



We know equation having 2 zeroes is of form,

k (x² - (sum of zeroes) x + product of zeroes)

For a polynomial having and is its zeros the equation becomes,

x² + p/q x + 1/q = 0

So here we get,

g(x) = qx² + px + 1

Given,

f(x) = x² – p (x + 1) – c

α and β are the zeros

So,

f(x) = x² – p (x + 1) – c

= x² – px – (p + c)

As

(α + 1)(β + 1) = αβ + α + β + 1

= – p – c + p + 1

= 1 – c

Value of a polynomial at a point is described as the value obtained by the polynomial at that point in time.

For Example:

Polynomial f(x) = x³ – 2(1)² + 7(1) – 8

Now the value of the polynomial at x = 1

So here we get;

= 1 – 2 + 7 – 8

= – 2

A zero or we can say the root of a polynomial function is a number that, when put in for the variable, makes the function equal to zero. We use the Rational Zero Theorem to find all the zeros of a polynomial function and the possible rational roots of a polynomial equation.

Given

f(x) = x² – p (x + 1) – c

α and β are the zeros

Then,

f(x) = x² – p (x + 1) – c

= x² – px – (p + c)

As

(α + 1)(β + 1) = αβ + α + β + 1

= – p – c + p + 1

= 1 – c

So, the value of c,

c = 1

Given;

f(x) = ax² + bx + c has no real zeroes, and a + b + c<0

Suppose a = – 1,

b = 1,

c = – 1

Then a + b + c = – 1,

b² – 4ac = – 3

Therefore it is possible that c is less tha zero.

Suppose c = 0

Then b² – 4ac = b² ≥ 0

So,

f(x) has at least one zero.

Therefore c cannot equal zero.

Suppose c > 0.

It must also be true that b² ≥0

Then,

b² – 4ac < 0 only if a > 0.

Therefore,

a + b + c < 0.

– b > a + c > 0

b² > (a + c)²

b² > a² + 2ac + c²

b² – 4ac > (a – c)² ≥ 0

As we know that the discriminant can’t be both greater than zero and less than zero,

So, C can’t be greater than zero.

Given,

Sum of zeroes (α + β) = −(1/2)

Product of zeroes (αβ) = −3
As we know that the quadratic polynomial with zeroes α, β is given by;

x² − (α + β)x + (αβ)
= x² − [−(1/2)]x + (−3)
= x² + (1/2)x – 3

As seen from the graph,

The parabola cuts the graph at two points on the positive x-axis.

Hence, both the roots are positive.

Now, for a polynomial, the sum of roots is given as:

α + β = -b/a

∴ the sum will be positive as the roots are positive.

Also, the product of roots = c/a has to be positive too.

⇒ a is positive.

Now, since a is positive, therefore for the sum of roots to be negative, b has to be negative.

⇒ a > 0, b< 0 & c > 0.

Therefore, option (a) is correct.

Given,

– 1/4 and 1 are the zeros

So,

k[(x + 1/4)(x – 1)]

k[x² – 1x + 1/4x – 1/4]

f(x) = k[x² – 3/4 x – 1/4]

Where k is any non-zero real number.

As seen from the graph,

The parabola cuts the graph at two points on the x axis.

One root is positive & one root is negative.

Now, for a polynomial, the sum of roots is given as:

α + β = -b/a

Also, the product of roots = c/a is positive.

Because c is positive,

⇒ a is negative & b is negative.

Therefore, option (b) is correct.

Given,

Let the quadratic polynomial be f(x) = x² – 4x + k
Product of the zeroes of the quadratic polynomial = 3
Now,

Product of the zeroes = constant term/coefficient of x²

⇒ 3 = k/1
∴ k = 3

Given,

The quadratic polynomial f(x) = kx² – 3x + 5
Now,

Let two zeroes be a and 1 – a

∵ Sum of zeroes = 1 and also 1 – a ˃ 0 ⇒ a ˂ 1

Therefore;

Sum of the zeroes = 3/k

⇒ a + (1 – a) = 3/k

⇒ 1 = 3/k

⇒ k = 3

Given;

Polynomial = ax³ –6x² + 11x – 6

Let a,b,c be the zeroes of f(x) = ax³ –6x² + 11x – 6

Product of the zeroes = – (constant term)/coefficient of x³

4 = – ( – 6)/a

4 = 6/a

4a = 6

a = 6/4

a = 3/2

In figure 2.17, the polynomial has 2 zeroes because the graph cuts x-axis at 2 points i.e. x = – 3 and x = – 1.

Given;

f(x) = x³ – 3px² + qx – r

Let a, b and c be the zeroes of the polynomial x³ – 3px² + qx – r

a + b + c = 3 p

ab + bc + ac = q

abc = r

a + b = 2b ⇒ b = p ⇒ a + c = 2p

b(a + c) + ac = q

⇒ 2p² + ac = q

⇒ ac = r/p

∴ 2p² + r/p = q

2p³ = pq – r

In figure 2.18, the polynomial has 3 real zeroes because the graph cuts axises at three points.

Given;

f(x) = 2x³ + 6x² – 4x + 9

Product of the two zeros = 3

a = 2,

b = 6

c = – 4

d = 9

Let the zeros of f(x) be p, q, r

Product of the zeroes = – d/a

p×q×r = – 9/2

3r = – 9/2 … (given p×q = 3)

r = – 3/2

Hence,

Third zero = – 3/2

The signs of a will be positive as a > 0 and,

The signs of b² – 4ac will be positive as b² – 4ac > 0

Given;

f(x) = ax³ + bx – c is divisible by g(x) = x² + bx + c

Now by division Method,

As f(x) is divisible by g(x), then remainder must be 0,

i.e.

(ab² – ac + b)x + c(ab – 1) = 0

⇒ (ab² – ac + b)x = 0 and c(ab – 1) = 0

⇒ ab² – ab + b = 0 (∵ x ≠ 0) and ab – 1 = 0 (∵ c ≠ 0)

⇒ ab – 1 = 0

⇒ ab = 1

Given: f(x) = ax³ + bx – c is divisible by g(x) = x² + bx + c

To find: The value of c.

Solution:

Now by division Method first we have to calculate the value of ab,

As f(x) is divisible by g(x), then remainder must be 0,

i.e.

⇒ (ab² – ac + b)x + c(ab – 1) = 0

⇒ (ab² – ac + b)x = 0 and c(ab – 1) = 0

⇒ ab² – ac + b = 0 (∵ x ≠ 0)

and ab – 1 = 0 (∵ c ≠ 0)

⇒ ab – 1 = 0

⇒ ab = 1

Now take;

ab² – ac + b = 0

⇒ (ab)b – ac + b = 0

Using ab = 1 we get;

⇒ 2b – ac = 0

⇒ c = 2b/a ... (1)

Multiply and divide RHS of (1) by b

⇒ c = 2b/a = 2b²/ab

So we get;

c = 2b²

b² – 4ac = 0, Two

The given quadratic equation touches the x-axis at only one point.

The root of the quadratic equation is equal and real because if the quadratic equation has two distinct roots, then the graph touches the x-axis at two points.

As we know that the roots are real and equal if the value of discriminant is zero,

So,

b² – 4ac = 0

Given;

f(x) = 5x² + 13x + k

Let suppose roots are R and 1/R

Product of the roots = C/R = R×1/R

R×1/R = k/5

1 = k/5

So,

Here k = 5

The y-intercept of the equation is c,

So, the signs of will be positive as c > 0

Given,

f(x) = ax³ + bx² + cx + d,

As we know,

α + β + y = b/a

αβ + βy + yα = – c/a …………….(i)

αβy = d/a ……………….. (ii)

Now,

Divide (i) by (ii)

We get

1/y + 1/α + 1/β = – c/d

The y-intercept of the equation is c,

So, the signs of will be negative as c < 0

In the figure 2.21,

The polynomial has 4 real zeroes because the graph cuts x-axis at four points.

f(x) = ax³ + bx² + cx + d,

Then the roots are : α , β and γ

Now, sum of roots:

α + β + γ = - b/a ….(i)

The product of zeroes,

α × β × γ = -d/a….(ii)

And,

αβ + βγ + γα = c/a…(iii)

Now,

(α + β + γ)² = α² + β² + γ² + 2αβ + 2βγ + 2γα

⇒

⇒

Given,

f(x) = x³ – 2x² + 4x + k

x = 1 is a zero,

So,

Keeping x = 1 we get,

f(1) = 1³ – 2(1)² + 4×1 + k

f(1) = 1 – 2 + 4 + k

f(1) = 5 – 2 + k

f(1) = 3 + k

– 3 = k

So we get,

k = – 3

Given,

f(x) = x³ – px² + qx – r

α, β and y are the Zeros,

α + β + y = – p

αβ + βy + yα = – q

αβy = – r

1/αβ + 1/βy + 1/yα = p/r

Given

f(x) = ax² + bx + c

α, β are the Zeros

α + β = – b/a

αβ = c/a

1/ α + 1/β =

1/α² + 1/β² = (1/α + 1/β)² – 2/αβ

= ( – b/c)² – 2a/c

=

The polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree; it is a generalized version of the familiar arithmetic technique called long division. It can be done manually because it separates a complex division problem into smaller ones.

Let’s take the Example:

f(x) and g(x) are two polynomials with,

g(x)≠0,

Now we can find the polynomials p(x) and q(x) such that,

f(x) = p(x) × g(x) × q(x)

Where q(x) = 0 or degree of q(x) < is degree of g(x).

The result is;

Dividend = Quotient × Divisor + Remainder

This is known as the Division Algorithm for polynomials.

Let

f(x) = x³ + x² + x + 1

g(x) = x + 2

q(x) = x² - x + 3

r (x) = -5

Now, f(x) = g(x) .q(x)+ r(x)

Here,

L.H.S = x³ + x² + x + 1

And, R.H.S = [(x + 2) × (x² - x + 3)] + (-5)

= x³ + 2x² – x² – 2x + 3x + 6 – 5

= x³ + x² – 2x + 3x + 6 – 5

= x³ + x² + x + 1

= R.H.S

∴ the given set of polynomials satisfies the equation: f(x) = g(x) .q(x)+ r(x)

Given,

ax³ + bx² + cx + d

By Putting x = 0

0 + d = 0

d = 0

ax³ + bx² + cx + d = 0

x(ax² + bx + c) = 0

Put x = 0

c = 0

ax² + bx = 0

x(ax + b) = 0

Hence,

x = – b/a

Given,

x³ + x² – 5x – 5

Zeros = √5 and – √5

x² – 5 is a root

(x² – 5)(x + 1) = 0

x = – 1

So,

The third zero is – 1

Given,

Sum of zeros = 2√3 and

Product = 2

As we know;

f(x) = x² + ( – Sum of zeros)x + (Product of zeros)

So,

f(x) = x² – 2√3x + 2

Given,

p(x) = x³ + 4x² + x – 6

As

a = 1

b = 4

c = 1

d = – 6

So the product of the zeros will be;

= – d/a

= – ( – 6)/1

= 6

Degree of reminder is less than the degree of divisor so the degree of reminder could be one or zero depending on the quadratic polynomial.

a = 2, b = 0

Given,

A polynomial f(x) = x³ + x² – ax + b

Which is divisible by x² – x

Now,

x² – x = (x – 1) = (x – 0)(x – 1)

(x – 0) and (x – 1) are factors of polynomial f(x)

⇒ f(0) = 0

⇒ 0³ + 0 – a × 0 + b = 0

⇒ b = 0

And f(1)

1³ + 1 – a × 1 + b = 0

⇒ 2 – a = 0

⇒ a = 2

∴ a = 2 and b = 0

Given;

p(x) = x² – 5x + 4

3 is the zero of the resulting polynomial

So we have;

p(3) = (3)² – 5×3 + 4

p(3) = 9 – 15 + 4

p(3) = – 2

As – 2 is the remainder,

So 2 should be added to the given polynomial to get 3 as the zero of the resulting polynomial.

We can also verify the answer;

By adding 2 to the given polynomial we get;

P(x) = x² – 5x + 4 + 2

P(x) = x² – 5x + 6

P(3) = 9 – 15 + 6

P(3) = 15 – 15

P(3) = 0

Hence proved that 2 is the right answer to get 3 as the zero of resulting polynomial.

Given;

A polynomial f(x) = 2x³ – 6x² + 5x – 7

And zeroes are a – b, a and a + b,

Let’s take a = α

b = β

c = y

As we know that,

α + β + y = – b/a

(a – b) + a + (a + b) = – ( – 6)/2

3a = 3

a = 1

So, the value of a = 1

Given,

p(x) = x² – 16x + 30

15 is the zero of the resulting polynomial,

So we get;

p(15) = (15)² – 16 × 15 + 30

p(15) = 225 – 240 + 30

p(15) = 15

∴ 15 should be subtracted from the given polynomial to get 15 as the zero of the resulting polynomial.

We can also verify the answer;

By adding 2 to the given polynomial we get;

P(x) = x² – 16x + 30 – 15

P(x) = x² – 16x + 15

P(15) = 225 – 240 + 15

P(15) = 240 – 240

P(15) = 0

Hence proved that 15 is the right answer to get 15 as the zero of resulting polynomial.

Given,

Polynomial p(z) = z⁵ – 2z² + 4

Now first break the given equation,

We get,

Coefficient of z⁵ = 1

Coefficient of z⁴ = 0

Coefficient of z³ = 0

Coefficient of z² = – 2

Coefficient of z = 0

Constant term = 4

Therefore 1, 0, 0, – 2, 0, 4 are coefficients of the polynomial p(z)

Given,

One of the polynomial is 3,

Sum of the zeros = 0

Now

Let the other zero be x,

x + 3 = 0

x = – 3

So,

Zeros of the polynomial are 3 and – 3

Product of the zeros = 3 × – 3 = – 9

Polynomial with zeros 3 and – 3 = x² – (0)x + ( – 9) = x² – 9

Given,

Polynomial = X² – x – 6 = 0
x² – 3x + 2x – 6 = 0
x(x – 3) + 2(x – 3) = 0
(x + 2)(x – 3) = 0
So – 2 and 3 are the zeros of the given polynomial

Given

x³ + x² – 9x – 9

Zeros = 3 and – 3

x = – 1

So its third zero is – 1

Given,

x³ + 3x² – 5x – 15

Zeros = √5 and – √5

So the third zero = – 3

Given,

2x² + 2ax + 5x + 10

Factor = (x + a)

Now by using factor theoram we get,

f( – a) = 2a² – 2a² – 5a + 10 = 0

– 5a + 10 = 0

a = 2

Therefore the value of a = 2.

Given;

A factor = x + 2

So,

x + 2 = 0

x = – 2

p(x) = x² + ax + 2b

p( – 2) = ( – 2)² + ( – 2)a + 2b

0 = 4 – 2a + 2b

– 4 = – 2 (a – b)

2 = a – b……………Equation (i)

4 = a + b …….. Equation (ii)

By Adding (i) and (ii) we get,

2 – a + b + 4 – a – b = – 2a + 2

2a = 2

So we get, a = 3 and b = 1

Given,

A polynomial x² – x – (2k + 2)

– 4 is a zero of the given polynomial.

As – 4 is the zero,so at x = -4, the value of the polynomial x² – x – (2k + 2) will be 0.

So we get,

⇒x² – x – (2k + 2) = 0

⇒( – 4)² – ( – 4) – (2k + 2) = 0

⇒16 + 4 – (2k + 2) = 0

⇒20 – (2k + 2) = 0

⇒– (2k + 2) = – 20

⇒2k + 2 = 20

⇒ 2k = 20 - 2

⇒2k = 18

⇒k = 18/2
= 9

Given,

p(x) = ax² – 3(a – 1) x – 1

Zero is 1

Now,

p(1) = a(1)² – 3(a – 1) ×1 – 1 = 0

p(1) = a – 3a + 3 – 1 = 0

p(1) = – 2a + 2 = 0

= – 2a = – 2

= a = – 2/ – 2 = 1

So the value of a = 1

Given,

Divisor = – x² + x – 1

Quotient = x – 2

Remainder = 3

Now we have to find out Dividend…

As we know that,

Dividend = Divisor × Quotient + Remainder

Dividend = ( – x² + x – 1)(x – 2) + 3

= – x³ + x² – x + 2x² – 2x + 2 + 3

= – x³ + 3x² – 3x + 5

So,

The required polynomial is – x³ + 3x² – 3x + 5

For any polynomial, f(x) = ax² + bx + c

The sum of zeroes, is given as -b/a

& the product of zeroes is given as c/a

Here, the sum of zeroes is given as -6 = -6/1

⇒

⇒ b = 6 & a = 1

Also, product of zeroes = c/a

⇒

⇒ c = -4

& a = 1

∴ the polynomial is f(x)= x² +6x-4

Given,

Polynomial 2y² + 7y + 5

Zeroes = a, 13

Here a = 2

b = 7

c = 5

Lets take α and β are two zero,

So sum of the zeroes will be,

α + β = – b/a = – 7/2

Product of the zeroes = α. β = c/a = 5/2

Now put the values,

α + β + αβ = (α + β) αβ

= – 7/2 + 5/2 = – 2/2 = – 1

Given,

Polynomial 2x² + x + k

Zero = 3

By putting x = 3

We get,

p(x) = 2x² + x + k = 0

p(3) = 2(3)² + 3 + k = 0

p(3) = 18 + 3 + k = 0

= 21 + k = 0

k = – 21

Hence the answer is – 21

Given,

Polynomial x² + 11x + k

Zero of the polynomial = – 3

As we have zero,

f(x) = x² + 11x + k = 0

By putting x = – 3

f( – 3) = – 3² + 11×3 + k = 0

f( – 3) = 9 + ( – 33) + k = 0

f( – 3) = – 24 + k = 0

f( – 3) = k = 24

So we have the value of k = 24

Given,

Polynomial 3x² + 4x + 2k

Zero of the polynomial = – 2

As we have zero,

f(x) = 3x² + 4x + 2k = 0

By putting x = – 2

f( – 2) = 3( – 2)² + 4 ( – 2) + 2k = 0

f( – 2) = 12 – 8 + 2k = 0

= 4 + 2k = 0

= 2k = – 4

k = – 4/2 = – 2

So we have the value of k = – 2

Given,

Quadratic polynomial f(x) is factorizable into linear distinct factors;

So,

Let f(x) = (x – a)(x – b), where a ≠ b

If a, b are the element of R,

Then f(x) must be having two real and distinct zeroes.

True

Lets take,

f(x) = x² – 4x + 4

= (x – 2)²

= [g(x)]² ………….. [g(x) = (x – 2) is a linear polynomial]

Zero of g(x) is 2,

So,

Zeroes of f(x) are 2 and 2.

So we can say that zeroes of f(x) are coincident.

True

Let a polynomial f(x) = x² + 9

f(x) = (x + 3i)(x – 3i)

So, the zeroes are always imaginary and not real.

Let’s first draw the figure,

As we can see in the figure the number of zeroes can be 1 or 3.

So the least numbers of zeroes lying between a and b is 1.

Positive

If quadratic polynomial ax² + bx + c cuts positive direction of y-axis, then it means that the value of y is positive when x = 0 [∵ at y-axis, x = 0]

Now

y = c at x = 0 and y is positive.

So, c is positive.

Negative

If quadratic polynomial ax² + bx + c cuts Negative direction of y-axis, then it means that the value of y is negative when x = 0 [∵ at y-axis, x = 0]

Now,

y = c at x = 0 and y is Negative.

So, c is Negative.