Question

varify that -4is a zero of the polynomial p(×)=2׳+6ײ+7×+60​

Answer 1

Step-by-step explanation:

P(x) = 2x^3 + 6x^2 + 7x + 60

Taking -4 in place of x

P(-4) = 2(-4)^3 + 6(-4)^2 +7(-4) + 60

= (-2×64) + (6×16) -28 + 60

= -128 + 96 - 28 + 60

= -128 + 128

= 0

Therefore, (-4) is a zero of the required polynomial.

HENCE PROVED.

Answer 2

0,  verified

Step-by-step explanation:

Given, $p(x)=2x^3+6x^2+7x+60$

Given value of x= -4

So,

$P(-4)=2\times(-4)^3+6\times(-4)^2+7(-4)+60\\=2\times-64+6\times16-28+60\\=-128+96-28+60\\=-156+156\\=0\\verified$