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What must be added to x3-3x2-12x+19 so that the result is exactly divisible by x2+x-6?

User Robert N
by
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1 Answer

2 votes

Answer:


x^3-3x^2-12x+19 is divisible by
x^2+x-6 when
2x+5 is added to it.

Explanation:

Let
p(x)=x^3-3x^2-12+19 &
q(x)=x^2+x-6

By division algorithm, when p(x) is divided by q(x), the remainder is a linear expression in x

So, let
r(x)=ax+b is added to
p(x) so that
p(x)+r(x) is divisible by
q(x)

Let,
f(x)=p(x)+r(x)


f(x)=x^3-3x^2-12x+19+ax+b


f(x)=x^3-3x^2+x(a-12)+b+19

We have,


q(x)=x^2+x-6=(x+3)(x-3)

Clearly,
q(x) is divisible by
(x-2) and
(x+3)

{
(x-2) and
(x+3) are factors of
q(x) }

We have,


f(x) is divisible by
q(x)


(x-2) and
(x+3) are factors of
f(x)

From factors theorem,

If
(x-2) and
(x+3) are factors of
f(x)

then
f(2)=0 and
f(-3)=0 respectively,


f(2)=0
2^3-3(2)^2+2(a-12)+b+19=0


8-12+2a-24+b+19=0


2a+b-9=0

Similarly,


f(-3)=0
(-3)^3-3(-3)^2+(-3)(a-12)+b+19=0


-27-27-3a+36+b+19=0


b-3a+1=0

Subtract (1) from (2)


b-3a+1-(2a+b-9)=0-0


b-3a+1-2a-6+9=0


-5a+10=0
5a=10
a=2

Put
a=2 in equation 2


b-3 ×
2+1=0
b-6+1=0
b-5=0
b=5


r(x)=ax+b
r(x)=2x+5

Hence,
x^3-3x^2-12x+19 is divisible by
x^2+x-6 when
2x+5 is added to it.

Hope this helps...

User Euge
by
8.1k points

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