give a complete factored form of the polynomial $-6a^5+48a^4+12a$

Question

Give a complete factored form of the polynomial $-6a^5+48a^4+12a$

I have tried solving this equation and I just cant figure it out. Help me, and give me the answer.

Answer 1

You can start by factoring out $-6a$,

$$-6a(a^4 - 8a^3 - 2)$$

The question now is, is $a^4 - 8a^3 - 2$ factorizable? To answer this, we use Rational Zero Theorem. Base on that theorem, the roots must have a numerator a factor of $d_0 = -2$ and denominator a factor of $d_n = d_4 = 1$, thus the root must be a factor of $-2$.

By testing all the factors of $-2$ (other than 0), $-1, 1, -2, 2$, we get $7, -9, 78, -50$ respectively. None of which is 0, thus none of the factors of $-2$ are root of $a^4 - 8^3 - 2$, hence $a^4 - 8^3 - 2$ can't be factored much further. Thus, the final factored form is,

$$-6a(a^4 - 8a^3 - 2)$$

Answer 2

When factoring any expression, it helps to think of what you're doing as working backwards through a multiplication problem. Remember that you're trying to come up with terms whose product equals the expression you have started with originally. Some polynomials can be easily recognized as a product of common factors, such as $n^2 -1$, which when completely factored is $(n+1)(n-1)$. However, when you aren't working with something recognizable, you want to look for common factors and remove them until you don't have anymore.

For your polynomial, you want to start with what you can recognize as common to all terms. If we factor out $-6a$, you're left with:

$$-6a^5+48a^4+12a \Rightarrow -6a(a^4-8a^3-2) $$

Now, we can stop there or we can continue. If we examine our new terms, we'll see the common factor $a^3$ in two of the terms. If we remove this, the polynomial becomes:

$$-6a(a^4-8a^3-2) \Rightarrow -6a(a^3(a-8)-2) $$

If you need to make sure you've factored correctly, remember to always multiply your factors together and make sure your product is the polynomial you started with.