Question

if the polynomials ax^3+3x^2-13 and 2x^3-5x+a when divided by (x – 2), leave the same remainders, find the value of a.

Answer 1

Here the SOLUTION goes

Zero of (x-2) =2

$let \: f(x) = a {x}^{3} + 3 {x}^{2} - 13$

$f(2) = a {(2)}^{3} + 3 {(2)}^{2} - 13$

$f(2) = 8a+ 12 - 13$

$f(2) = 8a - 1....(i)$

Again,

$let \: g(x) = 2 {x}^{3} - 5 {x} + a$

$g(2) = 2 {(2)}^{3} - 5 {(2)} + a$

$g(2) =16 - 10+ a$

$g(2) = 6 +a.......(ii)$

By QUESTION,

8a -1 = 6+ a

=> 8a - a = 6 + 1

=> 7a = 7

=> a = 7/7 = 1

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Answer 2

Answer:

Let f(x)=ax^3+3x^2-13 and g(x)=2x^3-5x+a

given divisor =x-2

x-2=0=>x=2 (BY FACTORS THEOREM)

given when the polynomials are divided by x-2

we will get the same remainder let that be R

f(2)=a(2^3)+3(2^2)-13 and g(2)=2(2^3)-5(2)+a

=>a(8)+3(4)-13 and 2(8)-5(2)+a

=>8a-1=R.........(1) and a+6=R.........(2)

(1)-(2)

=>8a-1-a-6=R-R

=>7a-7=0

=>7a=7

=>a=7