# Find the real zeros of the function algebraically

1. Aug 31, 2010

### swatmedic05

f(x)=x^3+5x^2+9x+45
x^3+5x^2+9x+45=0
x^2(x+5)+9(x-5)=0
(x^2+9)(x+5)=0
What happends to the (x+5)?

2. Sep 1, 2010

### Staff: Mentor

There's an error in the step above.
Since (x2 + 9)(x + 5) = 0, then either x2 + 9 = 0 or x + 5 = 0. Can you continue from here?

3. Sep 1, 2010

### eumyang

\begin{aligned} x^3 + 5x^2 + 9x + 45 &= 0 \\ x^2(x + 5) + 9(x + 5) &= 0 \\ (x^2 + 9)(x + 5) &= 0 \end{aligned}
I'm assuming that the error pointed out above is just a typo, so I'll just fix it. As for
It was the greatest common factor, so it got factored out.

Say you wanted to factor 16x - 28. The GCF is 4, and you could rewrite the expression as
4(4x) - 4(7),
and then factor out the 4 to get
4(4x - 7).

It's the same thing in your problem. The GCF is an expression: x + 5. You factor the x + 5 out in $$x^2 \bold{(x + 5)} + 9 \bold{(x + 5)} = 0$$ and you get $$(x^2 + 9)(x + 5) = 0$$. Maybe part of the confusion is that the GCF appears last instead of first?

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Last edited: Sep 1, 2010
4. Sep 1, 2010

### HallsofIvy

It was the greatest linear factor. It is not a monomial, it is a binomial, just like $x^2+ 9$.

5. Sep 1, 2010

### eumyang

Fixed. Didn't mean to use the term "monomial" -- just GCF. This is what you get for replying at 5am. :P

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