Find the value of a and b so that $$x ^{4}-ax ^{3}-\left( 6a+5b \right)x ^{2}+abx+144$$ can be divided by $$x ^{2}+6x+8$$!

I've tried to use long division but it seems doesn't work...,

Question

Find the value of a and b so that $$x ^{4}-ax ^{3}-\left( 6a+5b \right)x ^{2}+abx+144$$ can be divided by $$x ^{2}+6x+8$$!

I've tried to use long division but it seems doesn't work...,

hartnn · Answer

hint : if that expression can be divided by that quadratic equation, then roots of that quadratic equation satisfy that 4th degree expression.

hartnn · Answer

say if you get 2 roots as p and q , 
then put x=p, and x=q in that 4th degree equation, to get 2 equations in a and b which can be solved simultaneously ....

chihiroasleaf · Answer

the roots of $$x ^{2}+6x+8$$ are -2 and -4

what is f(-2) and f(-4) ?
are f(-2) and f(-4) equals zero?

chihiroasleaf · Answer

I mean the roots of quadratic equation $$x ^{2}+6x+8=0$$

anonymous · Answer

x^2 +6x+8 =(x+2) (x+4)
So,
Let f(x)= x^4 -ax^3 -(6a+5b)x^2+abx+144

hartnn · Answer

yes, because f(x) can be divided without remainder , so, f(-2)=0 , f(-4)=0.

anonymous · Answer

Since f(x) is divided by x^2+6x+8
f(-2)=0
f(-4)=0
U will get 2 equations 
and u need to solve them

parthkohli · Answer

The remainder theorem says that the remainder of $\dfrac{p(x)}{q(x)}$ is $p(a)$ where $a$ is the root of $q(x)$.

chihiroasleaf · Answer

I just know the remainder theorem for division by linear expression like (x-a)
so it also hold for quadratic expression?

hartnn · Answer

remember that quadratic expression is product of 2 linear expressions, so yes.

chihiroasleaf · Answer

hhmm..., if (x-a)(x-b) divide f(x) ...,
is (x-a) divide f(x) and (x-b) divide f(x) ?

if the remainder theorem holds, then it should be true, right? 

I'm confused with this.., so I'm not sure in applying remainder theorem before...

hartnn · Answer

example, if (x-4)(x+4) divide f(x) [let f(x)=x^2-16]
then isn't it true that both x-4 and x+4 divide x^2-16
and f(4) = f(-4) =0.

chihiroasleaf · Answer

I see.... :D

then, 
I get this..
f(-2) = 10a + 5b -ab = -80
f(-4) = 22a+5b+ab = -100

I add both equation, then I get 16a+5b = -90... 
and I get stuck... :(

chihiroasleaf · Answer

I think I've an idea to solve it..., 
thanks a lot.... :)

hartnn · Answer

u sure those equations are correct ? http://www.wolframalpha.com/input/?i=solve+10a+%2B+5b+-ab+%3D+-80+%2C+22a%2B5b%2Bab+%3D+-100

chihiroasleaf · Answer

ahh.., I made mistake in calculation  
it should be 
$$f(-2) = 10a+5b-ab=-80$$

$$f(-4)=  22a+5b-ab=-100$$

hartnn · Answer

then (-5/3 , -19/2)

chihiroasleaf · Answer

yess..., I get it..., 
thanks a lot... :D

hartnn · Answer

welcome ^_^