Question 1153478
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If two {{{cross(factors)}}} roots of the polynomial are 2 and -4, evaluate a and b
P(x) = x^3+ax^2+bx-8<br>
For an odd degree polynomial, the product of the roots is the opposite of the constant term, divided by the leading coefficient.  So for this polynomial, the product of the three roots is<br>
{{{(-(-8))/1= 8}}}<br>
The product of the two given roots is -8; that means the third root is -1.<br>
The coefficient of the x^2 term, a, is the opposite of the sum of the roots:<br>
{{{-((2)+(-4)+(-1)) = 3}}}<br>
The coefficient of the linear term, b, is the sum obtained by adding the products of the roots 2 at a time:<br>
{{{(2)(-4)+(2)(-1)+(-4)(-1) = -8-2+4 = -6}}}<br>
So the polynomial is<br>
{{{P(x) = x^3+3x^2-6x-8}}}<br>
A graph, showing the three roots at -4, -1, and 2:<br>
{{{graph(400,400,-5,5,-20,20,x^3+3x^2-6x-8)}}}<br>