Question

check whether x-1 is a factor of
${x}^{3} + 6 {x}^{2} - 2x - 5$
By trial method.

PLEASE HELP ME....​

Answer 1

$\huge{\bf{\underline{\purple{\mathrm{\underline{\red{Hello}{\blue{Mate}}}}}}}}$

$\huge{\bf{\underline{\purple{\mathrm{\bigstar{\red{ANS}{\blue{WER}}}}}}}}$

Given : p(x) = x^3 + 6x^2 - 2x -5

(x-1) is the factor

so,

Let , x - 1 = 0

x = 1

Now substitute the value of -1 in place of x

so, we get

$p(x) = {x}^{3} + 6 {x}^{2} - 2x - 5 \\ p(0) = (1 {)}^{3} + 6(1 {)}^{2} - 2(1) - 5 \\ = 1 + 6 - 2 - 5 \\ = 7 - 7 \\ = 0$

Therefore,

(x-1) is the factor of p(x)

Hope this is useful....

Stay home....Stay safe......