Question

If ${{x}^{140}}+2{{x}^{151}}+k$ is divisible by x+1, then the value of k is
[a] 1
[b] -3
[c] 2
[d] -2

Answer 1

Hint: Recall factor theorem. According to factor theorem x-a is a factor of p(x) if p(a) = 0. Use factor theorem and hence form an equation in k. Solve for k. The value of k is the value at which x+1 is a factor of p(x).

Complete step-by-step answer:
We have $p\left( x \right)=2{{x}^{151}}+{{x}^{140}}+k$
Since x+1 is a factor of p(x), we have
$p\left( x \right)=\left( x+1 \right)g\left( x \right)$, where g(x) is some polynomial in x.
Put x = -1, we get
$p\left( -1 \right)=\left( -1+1 \right)g\left( -1 \right)=0\times g\left( -1 \right)=0$
Hence x = -1 is a zero of the polynomial p(x).
Now, we have $p\left( x \right)=2{{x}^{151}}+{{x}^{140}}+k$
Substituting x = -1 in the expression of p(x), we get
$p\left( -1 \right)=2{{\left( -1 \right)}^{151}}+{{\left( -1 \right)}^{140}}+k$
We know that ${{\left( -1 \right)}^{x}}=\left\{ \begin{matrix}
   1\text{ if }x\text{ is even} \\
   -1\text{ if }x\text{ is odd} \\
\end{matrix} \right.$
Now since 151 is odd, we have ${{\left( -1 \right)}^{151}}=-1$ and since 140 is even, we have ${{\left( -1 \right)}^{140}}=1$
Hence, we have
$p\left( -1 \right)=2\left( -1 \right)+1+k=-2+1+k=k-1$
But p(-1) = 0
Hence, we have
$k-1=0$
Adding 1 on both sides of the equation, we get
$k=1$
Hence, the value of k at which x+1 is a factor p(x) is 1.
Hence option [a] is correct.

Note: [1] Do not explicitly divide p(x) by x+1 and find the remainder as it is going to be very long since the quotient will have approximately 150 terms. Hence the long division method will be very long and cumbersome.