Question

find the zeroes of the polynomial x^square + 8x +12 and verify rhe relations between the zeroes and the coeffcients​

Answer 1

Answer:

x^2+8x+12

x^2+2x+6x+12

x(x+2)+6(x+2)

(x+6)(x+2)

x=-2 and. x=-6

sum=-b/a=-8

Zeroes sum=-2+(-6)=-8

product=c/a=12

zeroes product=-2*-6=12

hence verified

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Answer 2

Answer:

$\large{\underline{\boxed{\sf (-6) \: and \: (-2)}}}$

Step-by-step explanation:

Given polynomial ;

x² + 8x + 12

We can factorise it by splitting the middle term. We need rwo numbers such that their product is 12, and the sum is 8.

Such two numbers can be 6 and 2.

Here we go ;

⇒ x² + 8x + 12

⇒ x² + (6 + 2)x + 12

⇒ x² + 6x + 2x + 12

⇒ x(x + 6) + 2(x + 6)

⇒ (x + 6)(x + 2)

By zero product rule ;

⇒ x = - 6 or x = - 2

Hence, the zeroes of the polynomial are (-6) and (-2).

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Verification ;

We know that -

Sum of zeroes = $\bf \dfrac{-(Coefficient \: of \: x)}{Coefficient \: of \: x^2}$

⇒ - 6 - 2 = $\sf \dfrac{-8}{1}$

⇒ -8 = - 8

Hence, it is verified.

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Again,

Product of zeroes = $\bf \dfrac{Constant \: term}{Coefficient \: of \: x^2}$

⇒ - 6 * (-2) = $\sf \dfrac{12}{1}$

⇒ 12 = 12

Hence, it is verified.

_______________________

∴ The zeroes of the quadratic polynomial x² + 8x + 12 is (-6) and (-2).