The question I have is 2x^2+mx-21=0.
The area I am stuck on is finding one of the roots, one root was given as -7, however I need to find the other one.
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Product of the roots, may be. – Claude Leibovici Nov 09 '16 at 06:46
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Any reaction is appreciated. – callculus42 Nov 09 '16 at 07:30
1 Answers
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Some advice
You know that $x=-7$ is a root. Thus
$f(-7)=2\cdot (-7)^2-m\cdot 7-21=0$
Solve for $m$. After you have evaluated the value ($m^*$), you conduct the polynomial division with the known root.
$(2x^2+m^*x-21):(x+7)=\ldots$
And finally you set the result of the polynomial division equal to $0$ and solve for $x$ to get the second root.
callculus42
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