Question 1169662
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Let p(x) = 4x^2+6x+32


If x-4 is a factor of p(x), then p(4) = 0
This is because x-4 = 0 leads to x = 4 as a root.


So let's see if plugging x = 4 leads to 0 as a result
p(x) = 4x^2+6x+32
p(4) = 4(4)^2+6(4)+32
p(4) = 120
We get a nonzero value, so we conclude x-4 is not a factor of 4x^2+6x+32.


note how each term of 4x^2+6x+32 is positive
4x^2 is positive
6x is positive
32 is positive
Plugging a positive x value into the equation leads to some positive result. We need some negatives in there to counterbalance the positives, to be able to get to 0. This is a fairly quick way to see that p(4) cannot possibly be 0.



Answer: x-4 is <u>not</u> a factor of 4x^2+6x+32
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