Question

Find the zeroes of the following polynomials by factorization

method and verify the relations between the zeroes and the coefficients of the

polynomials
2. 3x2 + 4x – 4​

Answer 1

Answer:

Given :-

• 3x² + 4x - 4

To Find :-

• What is the zeroes and verify the relationship between the zeroes and co-efficient.

Solution :-

Given equation :

$\mapsto \sf\bold{3x^2 + 4x - 4}$

$\implies \sf 3x^2 + 4x - 4 =\: 0$

$\implies \sf 3x^2 + (6 - 2)x - 4 =\: 0$

$\implies \sf 3x^2 + 6x - 2x - 4 =\: 0$

$\implies \sf 3x(x + 2) - 2(x + 2) =\: 0$

$\implies \sf (3x - 2) (x + 2) =\: 0$

$\implies \sf (3x - 2) =\: 0$

$\implies \sf 3x - 2 =\: 0$

$\implies \sf 3x =\: 2$

$\implies \sf\bold{\purple{x =\: \dfrac{2}{3}}}$

$\implies \sf (x + 2) =\: 0$

$\implies \sf\bold{\purple{x =\: - 2}}$

$\therefore$ The zeroes of the polynomial is 2/3 and - 2.

Hence, we get :

• $\sf \alpha =\: \dfrac{2}{3}$

• $\sf \beta =\: - 2$

$\rule{150}{2}$

VERIFY THE RELATIONSHIP BETWEEN THE ZEROES AND CO-EFFICIENT :-

$\bigstar \: \: \sf\bold{\green{Sum\: of\: two\: roots\: :-}}$

As we know that :

$\clubsuit$ Sum of roots Formula :

$\longmapsto \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}$

So, given equation :

$\mapsto \sf\bold{3x^2 + 4x - 4}$

where,

• a = 3
• b = 4
• c = - 4

According to the question by using the formula we get,

$\longrightarrow \sf \alpha + \beta =\: \dfrac{- b}{a}$

$\longrightarrow \sf \dfrac{2}{3} + (- 2) =\: \dfrac{- 4}{3}$

$\longrightarrow \sf \dfrac{2}{3} - 2 =\: \dfrac{- 4}{3}$

$\longrightarrow \sf \dfrac{2 - 6}{3} =\: \dfrac{- 4}{3}$

$\longrightarrow \sf\bold{\red{ \dfrac{- 4}{3} =\: \dfrac{- 4}{3}}}$

Hence, Verified.

Again,

$\bigstar\: \: \: \sf\bold{\green{Product\: of\: two\: roots\: :-}}$

As we know that :

$\longmapsto \sf\boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}$

So, given equation :

$\mapsto \sf\bold{3x^2 + 4x - 4}$

where,

• a = 3
• b = 4
• c = - 4

According to the question by using the formula we get,

$\longrightarrow \sf \alpha\beta =\: \dfrac{c}{a}$

$\longrightarrow \sf \dfrac{2}{3} \times (- 2) =\: \dfrac{- 4}{3}$

$\longrightarrow \sf \bold{\red{\dfrac{- 4}{3} =\: \dfrac{- 4}{3}}}$

Hence, Verified.