Answer:
[tex]x^3-3x^2-12x+19[/tex] is divisible by [tex]x^2+x-6[/tex] when [tex]2x+5[/tex] is added to it.
Step-by-step explanation:
Let [tex]p(x)=x^3-3x^2-12+19[/tex] & [tex]q(x)=x^2+x-6[/tex]
By division algorithm, when p(x) is divided by q(x), the remainder is a linear expression in x
So, let [tex]r(x)=ax+b[/tex] is added to [tex]p(x)[/tex] so that [tex]p(x)+r(x)[/tex] is divisible by [tex]q(x)[/tex]
Let, [tex]f(x)=p(x)+r(x)[/tex]
⇒ [tex]f(x)=x^3-3x^2-12x+19+ax+b[/tex]
⇒ [tex]f(x)=x^3-3x^2+x(a-12)+b+19[/tex]
We have,
[tex]q(x)=x^2+x-6=(x+3)(x-3)[/tex]
Clearly, [tex]q(x)[/tex] is divisible by [tex](x-2)[/tex] and [tex](x+3)[/tex]
{ [tex](x-2)[/tex] and [tex](x+3)[/tex] are factors of [tex]q(x)[/tex] }
We have,
[tex]f(x)[/tex] is divisible by [tex]q(x)[/tex]
⇒ [tex](x-2)[/tex] and [tex](x+3)[/tex] are factors of [tex]f(x)[/tex]
From factors theorem,
If [tex](x-2)[/tex] and [tex](x+3)[/tex] are factors of [tex]f(x)[/tex]
then [tex]f(2)=0[/tex] and [tex]f(-3)=0[/tex] respectively,
⇒ [tex]f(2)=0[/tex] ⇒ [tex]2^3-3(2)^2+2(a-12)+b+19=0[/tex]
⇒ [tex]8-12+2a-24+b+19=0[/tex]
⇒ [tex]2a+b-9=0[/tex]
Similarly,
[tex]f(-3)=0[/tex] ⇒ [tex](-3)^3-3(-3)^2+(-3)(a-12)+b+19=0[/tex]
⇒ [tex]-27-27-3a+36+b+19=0[/tex]
⇒ [tex]b-3a+1=0[/tex]
Subtract (1) from (2)
[tex]b-3a+1-(2a+b-9)=0-0[/tex]
⇒ [tex]b-3a+1-2a-6+9=0[/tex]
⇒ [tex]-5a+10=0[/tex] ⇒ [tex]5a=10[/tex] ⇒ [tex]a=2[/tex]
Put [tex]a=2[/tex] in equation 2
⇒ [tex]b-3[/tex] × [tex]2+1=0[/tex] ⇒ [tex]b-6+1=0[/tex] ⇒ [tex]b-5=0[/tex] ⇒ [tex]b=5[/tex]
∴ [tex]r(x)=ax+b[/tex] ⇒ [tex]r(x)=2x+5[/tex]
Hence, [tex]x^3-3x^2-12x+19[/tex] is divisible by [tex]x^2+x-6[/tex] when [tex]2x+5[/tex] is added to it.
Hope this helps...
