Question 1167586
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The information p(-1)=34 is not needed to solve the problem....<br>
Since x=3+i is a root, x=3-i is another root.  The quadratic factor corresponding to those two roots is {{{x^2-6x+10}}}<br>
Given the leading coefficient and constant term of the polynomial, we know the factorization of the polynomial is of the form<br>
{{{(x^2-6x+10)(3x^2+nx-2)}}}<br>
Performing that multiplication yields<br>
{{{3x^4+(n-18)x^3+(28-6n)x^2+(10n+12)x-20}}}<br>
We can solve for n knowing that the coefficient of the quadratic term is 34<br>
{{{28-6n = 34}}}
{{{6n = -6}}}
{{{n = -1}}}<br>
We can then find the values of a and b, which are the coefficients of the cubic and linear terms of the polynomial:<br>
a = n-18 = -19
b = 10n+12 = 2<br>
ANSWERS: a = -19; b = 2
{{{3x^2-19x^3+34x^2+2x-20 = (x^2-6x+10)(3x^2-x-2) = (x^2-6x+10)(3x+2)(x-1)}}}<br>