PWN p. 151 #8 question

In PWN the SAT Math Guide (4th Ed, first printing p. 151), Polynomials chapter question #8, you explain the answer by graphing on calculator. Could you explain the answer to this question without graphing, in particular, why A, C and D are true and why B is false? Many thanks!

Sure! First, recognize that g(x) is factorable:

    \begin{align*}g(x)&=x^2+3x-10\\g(x)&=(x-2)(x+5)\end{align*}

From there, you can see that g(x) has two real zeros (2 and –5) so you can eliminate C.

Then, recognize that g(x) is a factor of f(x) as choice A says:

    \begin{align*}f(x)&=3x^3+9x^2-30x\\f(x)&=3x(x^2+3x-10)\\f(x)&=3x\left(g(x)\right)\end{align*}

Obviously that confirms that choice A is true, so you can eliminate A. However, that also confirms that choice B is false! Remember how we already factored g(x)!

    \begin{align*}f(x)&=3x\left(g(x)\right)\\f(x)&=3x(x-2)(x+5)\end{align*}

f(x) is divisible by (x + 5), but definitely not by (x – 5). You know from the way the question is constructed that only one choice will be false, so once you’ve found it, you can stop working and start bubbling. 🙂

Since you asked, though, we can also show that D is true from what we’ve already figured out using substitution. We already said f(x)=3x\left(g(x)\right), so we can substitute thusly:

    \begin{align*}h(x)&=f(x)+2g(x)\\h(x)&=3x\left(g(x)\right)+2g(x)\end{align*}

We can factor a g(x) out of each of those terms!

    \begin{align*}h(x)&=3x\left(g(x)\right)+2g(x)\\h(x)&=\left(g(x)\right)(3x+2)\end{align*}

Therefore, choice D is also true.