So far in this lesson, you have used long division to divide polynomials by a binomial. If the binomial is a factor, the long division allows you to determine a second factor. The following example shows how this process can be used to fully factor a polynomial and then use the factored form to sketch the corresponding graph.
Example: Factor and then Sketch f(x) = 2x3 + 6x2 − 20x − 48
Try dividing f(x) by x − 1:
The remainder is not zero, so x − 1 is not a factor of 2x3 + 6x2 − 20x − 48. |
Try dividing f(x) by x + 4:
The remainder is zero, so x + 4 is a factor of 2x3 + 6x2 − 20x − 48. |
Now you can rewrite f(x) as f(x) = (x + 4)(2x2 − 2x − 12). But you can go further; 2x2 − 2x − 12 can also be factored:
2x2 − 2x − 12 = 2(x + 2)(x − 3)
The fully factored form of f(x) is f(x) = 2(x + 4)(x + 2)(x − 3).
You now have two equivalent forms of this function:
Expanded: f(x) = 2x3 + 6x2 − 20x − 48
Factored: f(x) = 2(x + 4)(x + 2)(x − 3)
f(x) is a cubic polynomial with a positive leading coefficient, so you know the graph starts in quadrant 3 and ends in quadrant 1. The y-intercept is −48. The equivalent factored form tells you there are x-intercepts at −4, −2, and 3.
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In the previous example the binomials x − 1 and x + 4 were tested to see whether or not they were factors of a cubic polynomial. It was a lot of work to do the long division just to discover that x − 1 was not a factor.
You need a better way to see whether or not a binomial is a factor—long division is far too much work for that task. Try This 2 will help you see patterns that can be used to reduce the effort for determining factors of a polynomial.
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