Would you please show an easy way to solve this:

a(x) = x^3 +3x^2 + 5x
b(x) = 5x^2 + 17x + 16

Polynomial a(x) and b(x) are defined above. Which of the polynomials below has a factor of 3x + 2 ?

(A) l(x) = a(x)+ b(x)
(B) m(x) = 3a(x) + b(x)
(C) n(x) = a(x) – 3b(x)
(D) p(x) = 2a(x) + 3b(x)

This is a ripoff of Official Test 4 Section 4 #25, but it’s a bad ripoff because the original question could be easily factored for a clever and quick solution, and this one can’t. If you get a question like this, you’ll be allowed to use your calculator, though, so the following grind-it-out solution shouldn’t actually take that long.

Note that if a polynomial has a factor of 3x + 2, then it will also have a zero at x=-\dfrac{2}{3}. So the cleverest thing I can think of to do is calculate a\left(-\dfrac{2}{3}\right) and b\left(-\dfrac{2}{3}\right), and then plug those into each answer choice until one gives you zero. Like so:

    \begin{align*}a\left(-\dfrac{2}{3}\right)&=\left(-\dfrac{2}{3}\right)^3+3\left(-\dfrac{2}{3}\right)^2+5\left(-\dfrac{2}{3}\right)=-\dfrac{62}{27}\\b\left(-\dfrac{2}{3}\right)&=5\left(-\dfrac{2}{3}\right)^2+17\left(-\dfrac{2}{3}\right)+16=\dfrac{62}{9}\end{align*}

Now try each answer choice to see which equals zero when a(x)=\dfrac{62}{27} and b(x)=\dfrac{62}{9}. You might be able to see it without much trouble…

A) -\dfrac{62}{27}+\dfrac{62}{9}=\dfrac{124}{27}

B) 3\left(-\dfrac{62}{27}\right)+\dfrac{62}{9}=0

C) -\dfrac{62}{27}-3\left(\dfrac{62}{9}\right)=-\dfrac{620}{27}

D) 2\left(-\dfrac{62}{27}\right)+3\left(\dfrac{62}{9}\right)=\dfrac{434}{27}

Obviously, the answer is B.

Another way to go is to graph each choice with your calculator and look for a zero at x=-\dfrac{2}{3}. That feels to me like it’ll be a bit time consuming, but if you just put your head down and started working on that the minute you saw this question, I bet you’d be done with it within 90 seconds, which is a fine amount of time to spend on a hard question in the calculator section.

 

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