Question

10. Using factor theorem, find the value of 'a' if the polynomial
p(x) = 2x4 - ax3 + 4x2 - x + 2 is divisible by (2x + 1).​

Answer 1

Given polynomial-

• 2x⁴ - ax³ + 4x² - x + 2

Factor-

• 2x + 1

To find-

• Value of a.

SoluTion-

According to question, given polynomial is divider by 2x + 1. So 2x + 1 is a factor of this polynomial.

2x + 1 = 0

→ x = -1/2

Put x = -1/2 in given polynomial,

2(-1/2)⁴ - a(-1/2)³ + 4(-1/2)² - (-1/2) + 2 = 0

→ 2(1/16) + a/8 + 1 + 1/2 + 2 = 0

→ 1/8 + a/8 + 1/2 + 3 = 0

→ (1 + 4 + 24)/8 + a/8 = 0

→ 29/8 + a/8 = 0

→ a/8 = -29/8

→ a = -29

Hence, value of a will be -29.

Answer 2

$\huge\mathtt\red{Question}$

Using factor theorem, find the value of 'a' if the polynomial

$\mathtt\orange{p(x)\: =\: 2\:x\:4 \:- \:a\:x\:3\: +\: 4\:x\:2 \:- \:x\: + \:2\: is\: divisible \:by\: (2x \:+ \:1).}$

$\huge\mathtt\red{To\:find}$

$\mathtt\longrightarrow\orange{Value\:of\:a.}$

$\mathtt\green{Factor\:is:}$

$\mathtt\longrightarrow\orange{2x\:+\:1}$

$\huge\mathtt\red{Solution}$

$\mathtt\orange{According\:to\:the\:question}$

$\mathtt\pink{2x\:+\:1}$

$\mathtt\longrightarrow\pink{x\:=\:-\frac{1}{2}}$

$\mathtt\orange{Putting\:the\:value\:of\:x}$

$\mathtt\pink{2(-\frac{1}{2}^4)-a(-\frac{1}{2}^3)+4(-\frac{1}{2}^2)-(-\frac{1}{2})+2=0}$

$\mathtt\longrightarrow\pink{2(\frac{1}{16})+\frac{a}{8}+1+\frac{1}{2}+2=0}$

$\mathtt\longrightarrow\pink{\frac{1}{8}+\frac{a}{8}+\frac{1}{2}+3=0}$

$\mathtt\longrightarrow\pink{\frac{1+4+24}{8}+\frac{a}{8}=0}$

$\mathtt\longrightarrow\pink{\frac{29}{8}+\frac{a}{8}=0}$

$\mathtt\longrightarrow\pink{\frac{a}{8}=-\frac{29}{8}}$

$\mathtt\longrightarrow\pink{a=-29}$

$\huge\mathtt\red{Answer}$

$\mathtt\orange{The\:value\:of\:a\:is\:-29.}$