Hi! On q8 page 149 (polynomials), what is happening in D? Why is it true?

This question (image below) asks which statement is FALSE. Generally, the best way to approach a question like this is to evaluate the simplest answer choices first and see if you can find one that’s false. If you do that, you probably never have to try D since B is much simpler to test and (spoiler alert) turns out to be false. But since you’re asking what’s going on in D, see below.

Choice D defines a new function, h(x). Let’s figure out what that looks like.

h(x)=f(x)+2g(x)\\h(x)=3x^3+9x^2-30x+2(x^2+3x-10)\\h(x)=3x^3+9x^2-30x+2x^2+6x-20\\h(x)=3x^3+11x^2-24x-20

Now, is that divisible by 3x+2? Well, if it were, then it should have a zero when 3x+2=0, right? Which means it’d have a zero at x=-\dfrac{2}{3}. Does it?

As it turns out, yes! You can see this quickly by graphing, or by plugging -\dfrac{2}{3} in for x throughout the equation:

h(x)=3x^3+11x^2-24x-20\\h(-\dfrac{2}{3})=3\left(-\dfrac{2}{3}\right)^3+11\left(-\dfrac{2}{3}\right)^2-24\left(-\dfrac{2}{3}\right)-20\\\\h(-\dfrac{2}{3})=3\left(-\dfrac{8}{27}\right)+11\left(\dfrac{4}{9}\right)-24\left(-\dfrac{2}{3}\right)-20\\\\h(-\dfrac{2}{3})=-\dfrac{8}{9}+\dfrac{44}{9}+\dfrac{48}{3}-20\\\\h(-\dfrac{2}{3})=\dfrac{36}{9}+\dfrac{48}{3}-20\\\\h(-\dfrac{2}{3})=4+16-20\\\\h(-\dfrac{2}{3})=0

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