Question 1090396: Use the p over q method and synthetic division to factor the polynomial P(x). Then solve P(x)=0
P(x)=x^4+11x^3+41x^2+61x+30
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The rational roots theorem tells us that the only possible rational roots are (plus or minus) 1, 2, 3, 5, 6, 10, 15, and 30. Furthermore, if you know Descartes' rule of signs, you know that the only possible rational roots are negative. So in the synthetic division process where you are testing for possible roots, you don't need to try any positive values.
We always try the smallest (in absolute value) roots first, both because most problems work that way, and because the calculations are easier.
I don't now how to show the synthetic division; but it turns out -1 is a root. The synthetic division shows
Occasionally in a problem like this the reduced polynomial can be factored by some familiar method, such as grouping. But such is not the case with this example.
Synthetic division shows -1 is not a root twice, so we try -2; and it too turns out to be a root. The synthetic division this time leaves us with the quadratic polynomial x^2+8x+15, which factors easily to(x+3)(x+5).
So
and so the solution of P(x)=0 is x = -1, -2, -3, or -5.
Your question specifically says to use the p/q method and synthetic division to factor the polynomial. However, there are other methods that are sometimes easier.
You might be familiar with rules about the sum and product of the roots of a polynomial that tell us that in your polynomial the product of the roots is 30 and the sum is -11. We know that the roots are all negative, and that the possible roots are -1, -2, -3, -5, -6, -10, -15, and -30.
And a bit of simple trial and error shows that the only combination of these roots (possibly used more than once each) that gives a product of 30 and a sum of -11 is -1, -2, -3, and -5.
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