If $x^2 + ax - b$ is a factor of $x^3 - 2bx^2 + ax -6$, show that $a = -2b -6/b$.
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Davidovich
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Try using polynomial long division. – player3236 Nov 06 '20 at 18:35
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I did and I get a = -2b but but can't seem to get the -6/b – Davidovich Nov 06 '20 at 18:36
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Can you show us what you did? You might have forgotten to carry the $-6$ term. – player3236 Nov 06 '20 at 18:40
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Through long division I get -2bx^2 - ax^2 + ax + bx - 6 = 0 – Davidovich Nov 06 '20 at 18:44
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I also tried to let the other factor equal (x - q) – Davidovich Nov 06 '20 at 18:44
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We know that $q(x^2+ax-b) = (-2b-a)x^2 + (a+b)x - 6$. Now we can compare coefficients. Also we do not have the $=0$ since this is not an equation. – player3236 Nov 06 '20 at 18:45
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this seems just the first step of dividing! you have to continue. – trula Nov 06 '20 at 18:48
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Not getting it. – Davidovich Nov 06 '20 at 18:54
1 Answers
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Suppose $x^3-2bx^2+ax-6$ has the factor $x^2+ax-b$. Suppose the other factor is $x-q$. Then:
$$(x-q)(x^2+ax+b)=x^3-2bx^2+ax-6$$
But by expanding:
$$(x-q)(x^2+ax+b)=x^3-qx^2+ax^2-qax+bx-qb=x^3+(a-q)x^2+(b-aq)x-bq$$
By comparing coefficients we have the equations:
$$a-q=-2b, \quad b-aq = a, \quad -bq=6$$
Hence $q=a+2b$ and $6 = -b(a+2b)$. This will reduce to $a = -2b-6/b$.
player3236
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