Introduction
Rational Roots Theorem
Review
Look at this quadratic equation: P(x) = x2 + 7x + 6
Do you recall the factors are (x + 6) and (x + 1)? And from that we get zeros of (−6) and (−1).
And if we evaluated P(−6) and P(−1) we would get zero for the remainder for each?
From the Fundamental Theorem of Algebra we know there are two roots/zeros and from the discriminant The discriminant is b2 − 4ac. a = 1, b = 7, c = 6. So, the discriminant is (7)2 − 4(1)(6). 49 − 24 = 25. 25 is positive we know they are real and there are two x-intercepts. Wow! Look at everything we have learned so far!
Finding Zeros
Let's use what we have learned to factor or find the zeros for
P(x) = x3 + 7x2 + x − 6
There is no GCF and it cannot be factored by grouping so we must try something else. I know there are three factors and I know that the three constant terms in the three factors should multiply together to give me −6. I also know that the remainder will equal zero for all the factors. What if I evaluate in P(x) until I find three zeros? I only need those positive and negative numbers that are a factor of −6.
Start with P(−6), then P(−3), P(−2), P(−1), P(1)...
Remember how to put the equation in the graphing calculator, if you have one, and let it do the work or use this evaluation tool.
Example #1 The Rational Roots Theorem
P(x) = x3 + 4x2 + x − 6
P(−6) = −84
P(−3) = 0
P(−2) = 0
P(−1) = −4
P(1) = 0
P(2) = 20
P(3) = 60
P(6) = 360
You should still have 3 zeros: {−3, −2, 1}.The three factors would be: (x + 3)(x +2)(x − 1)