Question

Find the zeroes of the following polynomials by factorisation method and verify the
relations between the zeroes and the coefficients of the polynomials:
2. 3x² + 4x - 4​

Answer 1

AnswEr

The zeroes of the polynomial are

2/3 and -2

Given

The quadratic polynomial :

• 3x² + 4x - 4

To Find

• The zeroes of the given polynomial and to verify the relations between the zeroes and the coefficients of the polynomial.

Solution

Factorisation of the polynomial by splitting the middle term :

3x² + 4x - 4

= 3x² + 6x - 2x - 4

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

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

Thus the zeroes of the polynomial are :

3x - 2= 0 and x + 2 =0

⇒x = 2/3 and ⇒x = -2

____________________

Now verification of the relationship

between the zeroes and the coefficients of the polynomial

3x² + 4x - 4

Here ,

•a = 3

•b = 4

•c = -4

We know that

→ Sum of the zeroes of the polynomial = -coefficient of x/coefficient of x²

⇒2/3+(-2) = -4/3

⇒2/3 - 2 = -4/3

⇒ (2 - 6)/3 = -4/3

⇒ -4/3 = -4/3 = -b/a

And again

→Product of the zeroes = constant term/coefficient ofx²

2/3×(-2) = -4/3

⇒ -4/3 = -4/3 = c/a

Thus Verified

Answer 2

Answer:

⋆ Given Polynomial : 3x² + 4x - 4

Here : a = 3,⠀b = 4,⠀c = - 4

$:\implies\tt f(x) = 0\\\\\\:\implies\tt 3x^2 +4x-4= 0\\\\\\:\implies\tt 3x^2+6x-2x-4= 0\\\\\\:\implies\tt 3x(x+2)-2(x+2) = 0\\\\\\:\implies\tt (3x-2)(x+2)=0\\\\\\:\implies\underline{\boxed{\tt x =\dfrac{2}{3}\quad or\quad x=-\:2}}$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{2}{3} + (-\:2) = \dfrac{-4}{3}\\\\\\\dashrightarrow\tt\:\: \dfrac{2-6}{3} = \dfrac{ - 4}{3} \\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt \dfrac{ - \:4}{3} = \dfrac{ -\:4}{3} }}}\\\\\\{\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{2}{3} \times (-\:2) = \dfrac{-4}{3}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt \dfrac{-\:4}{3} = \dfrac{-\:4}{3}}}}$