Question

19. If x – 1 is a factor of g(x) = x2 + kx + 1, then find k and hence prove that
(x - k) is a factor of
p(x) x3 + 3x2 + 3x + 2.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x - 1 \: is \: a \: factor \: of \: g(x) = {x}^{2} + kx + 1$

We know,

Factor theorem states that if a polynomial f (x) is divided by linear polynomial (x - a), then f(a) = 0.

$\rm :\implies\:g(1) = 0$

$\rm :\longmapsto\: {(1)}^{2} + k(1) + 1 = 0$

$\rm :\longmapsto\:1 + k + 1 = 0$

$\rm :\longmapsto\:k + 2 = 0$

$\bf\implies \:k = - 2$

Now,

Given that,

$\rm :\longmapsto\:p(x) = {x}^{3} + {3x}^{2} + 3x + 2$

Now, we have to show that x - k, i.e. x + 2 is a factor of p(x).

So, it is sufficient to show that p( - 2) = 0.

So,

Consider,

$\rm :\longmapsto\:p( - 2)$

$\rm  \:  =  \: \: {( - 2)}^{3} + 3 {( - 2)}^{2} + 3( - 2) + 2$

$\rm  \:  =  \: \: - 8 + 12 - 6 + 2$

$\rm  \:  =  \: \:14 - 14$

$\rm  \:  =  \: \:0$

$\bf\implies \:x + 2 \: is \: factor \: of \: p(x).$

Hence, Proved

Additional Information :-

Remainder Theorem :-

If a polynomial f(x) is divided by linear polynomial (x - a), then remainder is f(a).