Find the quadratic polynomial, sum of whose roots is 2 and their product is 1 3 .

We can find the quadratic equation if we know the sum of the roots and product of the roots. By using the formula f ( x ) = x 2 - ( α + β ) x + α β , where α , β are the roots of the quadratic polynomial. Given : Sum of the roots α + β = 2 and product of the roots α β = 1 3 ∴ f ( x ) = x 2 - 2 x + 1 3 Hence the quadratic polynomial is x 2 - 2 x + 1 3 .

Find the quadratic polynomial, sum of whose roots is 2 and their product is 1 3 .

We can find the quadratic equation if we know the sum of the roots and product of the roots. By using the formula f ( x ) = x 2 - ( α + β ) x + α β , where α , β are the roots of the quadratic polynomial. Given : Sum of the roots α + β = 2 and product of the roots α β = 1 3 ∴ f ( x ) = x 2 - 2 x + 1 3 Hence the quadratic polynomial is x 2 - 2 x + 1 3 .

One zero of the polynomial 3 x 3 + 16 x 2 + 15 x - 18 is 2 3 . Find the other zeros of the polynomial.

Given: x = 2 3 is one of the zero of 3 x 3 + 16 x 2 + 15 x – 18 . ⇒ x = 2 3 ⇒ x - 2 3 = 0 ⇒ 3 x - 2 = 0 Now, we divide 3 x 3 + 16 x 2 + 15 x – 18 by 3 x - 2 by long division to find the quotient, x 2 + 6 x + 9 3 x - 2 3 x 2 + 16 x 2 + 15 x - 18 3 x 2 - 2 x 2 - + 18 x 2 + 15 x - 18 18 x 2 - 12 x - + 27 x - 18 27 x - 18 0 0 So, the quotient is x 2 + 6 x + 9 . It means this quotient is another factor of the given polynomial. So, x 2 + 6 x + 9 = 0 and we further factorise this obtained factor. ⇒ x 2 + 3 x + 3 x + 9 = 0 ⇒ x ( x + 3 ) + 3 ( x + 3 ) = 0 ⇒ ( x + 3 ) ( x + 3 ) = 0 ⇒ ( x + 3 ) = 0 or ( x + 3 ) = 0 ⇒ x = – 3 o r x = – 3 Hence, the other zeroes of the polynomial are - 3 , - 3 & 2 3 .

Find a cubic polynomial whose zeros are 1 2 , 1 and - 3 .

If the zeros of the cubic polynomial are a , b and c then the cubic polynomial can be found as x 3 – ( a + b + c ) x 2 + ( a b + b c + c a ) x – a b c Let a = 1 2 , b = 1 and c = - 3 Substituting the values in above general equation, we get x 3 - ( 1 2 + 1 - 3 ) x 2 + ( 1 2 - 3 - 3 2 ) x - ( - 3 2 ) ⇒ x 3 - ( - 3 2 ) x 2 - 4 x + 3 2 ⇒ 2 x 3 + 3 x 2 - 8 x + 3 Hence, the required cubic polynomial is 2 x 3 + 3 x 2 - 8 x + 3

E M B I B E

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2A>Q 18

EASY

10th CBSE

IMPORTANT

Earn 100

Find the quadratic polynomial, sum of whose roots is $\sqrt{2}$ and their product is $\frac{1}{3}$ .

Important Questions on Polynomials

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2A>Q 19

If

x = \frac{2}{3}

and

x = - 3

are the roots of the quadratic equation

a x^{2} + 7 x + b = 0

, then find the value of

a

and

b

.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2A>Q 20

If

(x + a)

is a factor of the polynomial

2 x^{2} + 2 a x + 5 x + 10

, find the value of

a

.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2A>Q 21

One zero of the polynomial

3 x^{3} + 16 x^{2} + 15 x - 18

is

\frac{2}{3}

. Find the other zeros of the polynomial.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2B>Q 1

Verify that

3, - 2, 1

are the zeros of the cubic polynomial

p (x) = (x^{3} - 2 x^{2} - 5 x + 6)

and verify the relation between its zeros and coefficients.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>Exercise 2B>Q 2

Verify that $5, - 2$ and $\frac{1}{3}$ are the zeroes of the cubic polynomial $p (x) = (3 x^{3} - 10 x^{2} - 27 x + 10)$ and verify the relation between its zeroes and coefficients.