Question

If p( x ) = x¹⁰⁰ + ax + k is divisible by ( x² - x ) then find the value of a.

Answer 1

ANSWER:

Given:

• p(x) = x¹⁰⁰ + ax + k
• (x² - x) is a factor of p(x)

To Find:

• Value of a

Solution:

We are given that,

$\implies p(x)= x^{100}+ax+k$

We are also given that,

$\implies p(x^2-x)=0$

We can rewrite (x² - x) as,

$\implies x^2-x=x(x-1)$

So, the given polynomial is divisible by both x and (x - 1).

Hence, x and (x - 1) are both factors of p(x).

So,

$\implies x=0\:\:,\:x-1=0$

So,

$\implies x=0\:\:,\:x=1$

Hence,

$\implies p(0)=0\:\:,\:p(1)=0$

Taking 1st equation first,

$\implies p(0)=0$

$\implies 0= (0)^{100}+a(0)+k$

$\implies 0= 0+k$

Therefore,

$\implies\bf k=0$

Now,

$\implies p(x)= x^{100}+ax+k$

$\implies p(x)= x^{100}+ax+0$

$\implies p(x)= x^{100}+ax$

We had,

$\implies p(1)=0$

So,

$\implies p(1)= (1)^{100}+a(1)$

$\implies 0= 1+a$

$\implies\bf a=-1$

Therefore, value of a is -1.

Answer 2

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

• p( x ) = x¹⁰⁰ + ax + k is divisible by ( x² - x )

$\large\underline{\sf{To\:Find - }}$

• The value of a.

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

Given that,

$\rm :\longmapsto\:p(x) = {x}^{100} + ax + k$

Let assume that

$\rm :\longmapsto\:g(x) = {x}^{2} - x$

can be rewritten as

$\rm :\longmapsto\:g(x) = x(x - 1)$

It is given that p(x) = x¹⁰⁰ + ax + k is divisible by g(x) = ( x² - x ).

It implies that p(x) is divisible by x as well as p(x) is divisible by (x - 1).

We know,

By factor theorem, this theorem states that if a polynomial f(x) is divisible by linear polynomial (x - a), then f(a) = 0.

Now,

Case :- 1

When p( x ) = x¹⁰⁰ + ax + k is divisible by x.

By factor theorem,

$\rm :\longmapsto\:p(0) = 0$

$\rm :\longmapsto\:{(0)}^{100} + a(0) + k = 0$

$\bf :\longmapsto\: k = 0$

Case :- 2

When p( x ) = x¹⁰⁰ + ax + k is divisible by x - 1.

So, By factor theorem,

$\rm :\longmapsto\:p(1) = 0$

$\rm :\longmapsto\:{(1)}^{100} + a(1) + k = 0$

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

$\rm :\longmapsto\:1+ a + 0 = 0 \: \: \: \: \: \: \: \{ \: \because \: k \: = \: 0 \: \}$

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

$\bf :\longmapsto\:a = - 1$

Hence,

• The value of a for which p( x ) = x¹⁰⁰ + ax + k is divisible by ( x² - x ) is - 1.

Additional Information :-

Remainder Theorem :-

This theorem states that when a polynomial f(x) is divided by linear polynomial x - a, then remainder is f (a).