Factor P(x)=x3+6x2+11x+12 into linear and irreducible quadratic factors with real coefficients. Let P(x)=(x+a)(x2+bx+c). Then a=? b=? c=?

Answer 1

Let's use the method of synthetic division:

____1…6…11…12

-4| -4 -8 -12

___1....2.....3.......0

The first row of the table are the coefficients of P. We bring the 1 down to the last row (under the horizontal bar) and compute (-4)(1) = -4. Then we place the result -4 under 6 in the second column. Add 6 to -4 to obtain 2, which goes under the bar. Compute (-4)(2) = -8 and place -8 under the 11 in the third column. Do the same procedure to complete the last column. Since 0 is obtained in last entry under the bar, we know that not only if -4 a root of P(x), but in fact, P(x) = (x + 4)(x² + 2x + 3). So a = 4, b = 2, and c = 3.