$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
 A: by using the long division we get
$$\frac{ax^3+8x^2+bx+6}{x^2-2x-3}=ax+(8+2a)+\frac{(7a+b+16)x+6a+30}{x^2-2x-3}$$
now the reminder should be zero
$$7a+b+16=0\tag 1$$
$$6a+30=0\tag2$$
$$a=-5, b=19$$
A: What are the roots of $x^2-2x-3$? They should also be roots in the 3rd degree polynomial.
A: Suppose that $$ax^3+8x^2+bx+6=(x^2-2x-3)(ax-2).$$ And 
$$(x^2-2x-3)(ax-2)=ax^3-(2a+2)x^2+(4-3a)x+6.$$
So clearly $a=-5, b=19$. 
A: Let $f(x)=ax^3+8x^2+bx+6$, and let $g(x)=x^2-2x-3$. Since the question says that $g(x)|f(x)$, then write $f(x)=g(x)(px+q)$. Note the linear polynomial, we did that because dividing $f(x)$ by $g(x)$ will result in degree $1$. Now, expand the LHS and RHS, and you get something like this:
$$ax^3+8x^2+bx+6=(x^2-2x-3)(px+q)=px^3+x^2(q-2p)-x(2q+3p)-3q$$
Clearly, $q=-2,$ and $p=-5$. Now get the values of $a$ and $b$ by comparing the co-efficients. 
A: $$ x^2 - 2 x + 3 = ( x + 1) ( x -3) $$
$ x= -1, x = 3 $ should satisfy the polynomial. So,
$$ -a + 8 -b + 6 =0 ;\,  a \,3^3 + 8 \,3^2 - 3 \,3 + 6 =0 ;$$
Siimplify and solve for $a,b. $
A: if we divide both polynomials we get
$$xa+2\,a+8+1/4\,{\frac {27\,a+3\,b+78}{x-3}}+1/4\,{\frac {a+b-14}{x+1}}$$
and the remainder must be zero. And it must be
$$27a+3b+78=0$$ and $$a+b-14=0$$
A: I'll give you a hint: it's not the only way to do this, but it should be helpful.
Divide the first polynomial by the second, and you'll get a remainder as a first-degree polynomial $p(x)$ , with coefficients $f(a, b)$ and $g(a, b)$. Then, since you want the first polynomial to be divisible by the second, you want the remainder to be zero, so solve the system of equations $f(a, b)=0, \space g(a, b)=0$.
A: $$ax^3+8x^2+bx+6=(ax+c) (x^2-2x-3)=ax^3+(c-2a)x^2+(-2c-3a)x-3c$$
$$\begin{align}6&=-3c\Rightarrow c=-2\\8&=c-2a\Rightarrow 2a=c-8=-10\Rightarrow \color{red}{a=-5}\\b&=-2c-3a=4+15=\color{red}{19}\end{align}$$
Using this approach, you will also find out the third root is given by $ax+c=0\Rightarrow x=-\frac{2}{5}$
Also $x^2-2x-3x=(x-3)(x+1)$ gives $x=3$ and $x=-1$.
A: Hint: roots of one of them are roots of the other. This will give you a couple of equations which you can solve. 
