Question

Find the zeros of the polynomial 6x square + X - 2 and verify also

Answer 1

Step-by-step explanation:

6x^2+x-2=0

6x^2+4x-3x-2=0

2x(3x+2)-1(3x+2)=0

(3x+2)(2x-1)=0

(3x+2)=0 or (2x-1)=0

x=-2/3 or x=1/2.

Answer 2

$\huge\underline\mathtt\blue{Question}$

• Find the zeros of the polynomial 6x² + x - 2 and verify also it

$\huge\underline\mathtt\blue{To\:Find}$

• Zeros of polynomial
• Zeros of polynomialAlso verify it

$\huge\underline\mathtt\blue{Solution}$

Solve this question by splitting middle term

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

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

$\implies\sf 2x(3x+2)-1(3x-2)=0$

$\implies\sf (3x+2)(2x-1)=0$

So,

Either

$\implies\sf 3x+2=0$

$\implies\sf 3x=-2$

$\implies\sf x=\frac{-2}{3}$

or

$\implies\sf 2x-1=0$

$\implies\sf 2x=1$

$\implies\sf x=\frac{1}{2}$

-2/3 and 1/2 are the zeros of polynomial

$\huge\underline\mathtt\blue{Verification}$

$\large{\boxed{\bf{\orange{Sum\:of\:zeros}}}}$

Adding both zeros

$\sf =\frac{-2}{3}+\frac{1}{2}$

$\sf =\frac{-4+3}{6}$

$\sf =\frac{-1}{6}$

$\sf =\large\frac{-Coefficient\:of\:x}{coefficient\:of\:x^2}$

$\large{\boxed{\bf{\orange{Product\:of\:zeros}}}}$

Multiplying both zeros

$\sf =\frac{-2}{3}\times\frac{1}{2}$

$\sf =\frac{-2}{6}$

$\sf =\large\frac{Constant\:term}{coefficient\:of\:x^2}$