Question

2+4x" align="absmiddle" class="latex-formula">
$g(x)=x^2+3x+2$


The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by $2x+3$?


A) $h(x) = f(x) + g(x)$

B) $p(x) = f(x) + 3g(x)$

C) $r(x) = 2f(x) + 3g(x)$

D) $s(x) = 3f(x) + 2g(x)$

Answer 1

Answer:

B

Step-by-step explanation:

We have the two functions:

$f(x)=2x^3+6x^2+4x\text{ and } \\g(x)=x^2+3x+2$

First, let's factor each of the functions:

$f(x)=2x^3+6x^2+4x$

Factor out a 2x:

$f(x)=2x(x^2+3x+2)$

Factor:

$f(x)=2x(x+2)(x+1)$

Factor our second function:

$g(x)=x^2+3x+2$

Factor:

$g(x)=(x+2)(x+1)$

Now, let's see what happens if we add them together. This will yield:

$f(x)+g(x)=2x(x+2)(x+1)+(x+2)(x+1)$

We can rewrite this as:

$f(x)+g(x)=2x((x+2)(x+1))+1((x+2)(x+1))$

Using grouping (or the reverse of our distribute property), we can write this as:

$f(x)+g(x)=(2x+1)((x+2)(x+1))$

However, we want our function to be divisible by (2x+3). We have (2x+1).

So, we can multiply our g(x) by 3. If will go back, this will yield:

$f(x)+3g(x)=2x((x+2)(x+1))+3((x+2)(x+1))$

Now, we can do grouping again:

$f(x)+3g(x)=(2x+3)(x+2)(x+1)$

Since this now has a (2x+3) term, this is divisible by (2x+3).

So, the equation that is divisible by (2x+3) is f(x)+3g(x).

Our answer is B.

And we're done!