Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
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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 $$7a+b+16=0\tag 1$$ $$6a+30=0\tag2$$ $$a=-5, b=19$$
E.H.E
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What are the roots of $x^2-2x-3$? They should also be roots in the 3rd degree polynomial.
Olivier Moschetta
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I'm not sure, the question is word for word of what is in the title – Bill Smith Sep 17 '16 at 13:04
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1I'm giving you a hint. You should be able to find the roots of $x^2-2x-3$ yourself. It's a quadratic equation. – Olivier Moschetta Sep 17 '16 at 13:05
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(x+1)(x-3) but I don't know what to do next – Bill Smith Sep 17 '16 at 13:06
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So what are the roots? – Olivier Moschetta Sep 17 '16 at 13:07
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x=-1, x=3, I'm sorry we just started to learn polynomials so I'm struggling – Bill Smith Sep 17 '16 at 13:08
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No prob. If $P(x)=ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$, it means that the two roots $x=-1$ og $x=3$ are also roots in $P(x)$. And if $x=-1$ is a root then $P(-1)=0$. Can you finish the calculations? – Olivier Moschetta Sep 17 '16 at 13:09
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I got up to a + b -14 = 0 – Bill Smith Sep 17 '16 at 13:13
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Ok then use that $P(3)=0$ you should get a second equation for $a$ and $b$. Solve them together as a linear system. – Olivier Moschetta Sep 17 '16 at 13:15
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So I got P(3)= a(3)^3 + 8(3)^2 + b(3) + 6 , so a(27) + 72 + b(3) + 6, but I dont know what to do from here – Bill Smith Sep 17 '16 at 13:20
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You have two equations $a+b-14=0$ and $27a+3b+78=0$. This is a typical example of a linear system. You can for instance use the first equation to obtain $a=14-b$, plug it in the second equation and solve for $b$. – Olivier Moschetta Sep 17 '16 at 13:23
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So I got a= 5, b=19? – Bill Smith Sep 17 '16 at 13:27
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You should get $a=14-b=14-19=-5$, otherwise this is correct! – Olivier Moschetta Sep 17 '16 at 13:29
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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 $a=-5, b=19$.
Paul
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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.
codetalker
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$$ x^2 - 2 x + 3 = ( x + 1) ( x -3) $$
$ 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. $
Narasimham
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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 $$a+b-14=0$$
Dr. Sonnhard Graubner
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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 $f(a, b)=0, \space g(a, b)=0$.
Matija Sreckovic
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$$ax^3+8x^2+bx+6=(ax+c) (x^2-2x-3)=ax^3+(c-2a)x^2+(-2c-3a)x-3c$$ $$\begin{align}6&=-3c\Rightarrow c=-2\\8&=c-2a\Rightarrow 2a=c-8=-10\Rightarrow \color{red}{a=-5}\\b&=-2c-3a=4+15=\color{red}{19}\end{align}$$
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 $x=-1$.
msm
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Hint: roots of one of them are roots of the other. This will give you a couple of equations which you can solve.
Cameron Williams
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