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The polynomial ax3+bx2+x−6ax^3 + bx^2 + x - 6ax3+bx2+x−6​ has (x + 2) as a factor and leaves remainder 4 when divided by (x - 2). Find the value
Question

The polynomial ax3+bx2+x6ax^3 + bx^2 + x - 6​ has (x + 2) as a factor and leaves remainder 4 when divided by (x - 2). Find the value of a and b.

A.

a = 2, b = 0

B.

a = 1, b = 1

C.

a = 0, b = 2

D.

a = 7, b = 5

Correct option is C

Given:
f(x)=ax3+bx2+x6f(x) = ax^3 + bx^2 + x - 6 \\​​
Given conditions:
(i) (x+2)\text{(i) } (x + 2)​  is a factor =>f(2)=0 \Rightarrow f(-2) = 0 \\​​
(ii)\text(ii) ​Remainder when divided by (x2) (x - 2)​ is 4 =>f(2)=4 \Rightarrow f(2) = 4 \\[10pt]​​
Use f(2)=0f(-2) = 0 \\​​
f(2)=a(2)3+b(2)2+(2)6=0=>8a+4b26=0=>8a+4b=8(.......1)f(-2) = a(-2)^3 + b(-2)^2 + (-2) - 6 = 0 \\\Rightarrow -8a + 4b - 2 - 6 = 0 \\\Rightarrow -8a + 4b = 8 \quad \text{(.......1)} \\[10pt]​​
Use f(2)=4 f(2) = 4 \\​​
f(2)=a(2)3+b(2)2+26=4=>8a+4b4=4=>8a+4b=8(.......2)f(2) = a(2)^3 + b(2)^2 + 2 - 6 = 4 \\\Rightarrow 8a + 4b - 4 = 4 \\\Rightarrow 8a + 4b = 8 \quad \text{(.......2)} \\[10pt]​​
Equation 1: 8a+4b=8 -8a + 4b = 8 \\​​
Equation 2:   8a+4b=8 \ \ 8a + 4b = 8 \\​​
Add both equations:
(8a+4b)+(8a+4b)=8+88b=16=>b=2(-8a + 4b) + (8a + 4b) = 8 + 8 \\8b = 16 \Rightarrow b = 2 \\​​
Substitute  b = 2 into Equation 1:
8a+4(2)=8=>8a+8=8=>8a=0=>a=0-8a + 4(2) = 8 \Rightarrow -8a + 8 = 8 \Rightarrow -8a = 0 \Rightarrow a = 0 \\[10pt]​​
Correct answer is a=0, b=2a = 0,\ b = 2 \\[12pt]​​​​

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