Which of the binomials below is a factor of this trinomial? -2x^2 - 16x + 40 A. x - 5 B. x + 10 C. x - 10 D. x + 5
Take a polynomial function f(x) of degree 2 or more. If a binomial x - a is a factor of f(x), then f(a) = 0.
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Now look at the options. For binomial x - 5, a = 5 For binomial x + 10, a = -10 Do the same for the other two options. Then evaluate the polynomial at each a. If the polynomial evaluates to zero, that is the factor.
Here is another way of explaining this. Look at each option and compare it with x - a.
i still do not understand
A: x - 5 compare to x - a; a = 5 B: x + 10 compared to x - a: a = -10 C: x - 10 compared to x - a: x = 10 D: x + 5 compared to x - a: a = -5 Do you understand it so far?
not really
You need to have a binomial in the form x - a, where a number.
You must have x followed by a subtraction, then a number.
Look at option A: x - 5 Since you already have x followed by a subtraction, then the number 5, then for option A, which is x - 5, you can see that a = 5.
Option B has x + 10. In this case, x is followed by an addition, but we need the form x - a, so we write x + 10 as x - (-10) This hows that in this case, a = -10
We need each option to be in the form x - a A. x - 5 = x - a, so a = 5 B. x + 10 = x - (-10) = x - a, so a = -10 C. x - 10 = x - a, so a = 10 D. x + 5 = x - (-5) = x - a, so a = -5
Now we evaluate the given polynomial at each value of a. If it evaluates to zero, then that is the option that is a factor.
A. a = 5 -2x^2 - 16x + 40 = -2(5)^2 - 16(5) + 40 = -50 - 80 + 40 = -90 It does not evaluate to zero, so this is not it. B. a = -10 -2x^2 - 16x + 40 = -2(-10)^2 - 16(-10) + 40 = -200 + 160 + 40 = 0 It does evaluate to zero, so this is it. C. a = 10 -2x^2 - 16x + 40 = -2(10)^2 - 16(10) + 40 = -200 - 160 + 40 = -320 It does not evaluate to zero, so this is not it. D. a = -5 -2x^2 - 16x + 40 = -2(-5)^2 - 16(-5) + 40 = -50 + 80 + 40 = 70 It does not evaluate to zero, so this is not it.
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