Find the real zeros of the function algebraically

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swatmedic05
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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)?
 
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swatmedic05 said:
f(x)=x^3+5x^2+9x+45
x^3+5x^2+9x+45=0
x^2(x+5)+9(x-5)=0
There's an error in the step above.
swatmedic05 said:
(x^2+9)(x+5)=0
What happends to the (x+5)?

Since (x2 + 9)(x + 5) = 0, then either x2 + 9 = 0 or x + 5 = 0. Can you continue from here?
 
[tex]\begin{aligned}<br /> x^3 + 5x^2 + 9x + 45 &= 0 \\<br /> x^2(x + 5) + 9(x + 5) &= 0 \\<br /> (x^2 + 9)(x + 5) &= 0<br /> \end{aligned}[/tex]
I'm assuming that the error pointed out above is just a typo, so I'll just fix it. As for
What happened to the (x+5)?
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 [tex]x^2 \bold{(x + 5)} + 9 \bold{(x + 5)} = 0[/tex] and you get [tex](x^2 + 9)(x + 5) = 0[/tex]. Maybe part of the confusion is that the GCF appears last instead of first?69
 
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eumyang said:
[tex]\begin{aligned}<br /> x^3 + 5x^2 + 9x + 45 &= 0 \\<br /> x^2(x + 5) + 9(x + 5) &= 0 \\<br /> (x^2 + 9)(x + 5) &= 0<br /> \end{aligned}[/tex]
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 monomial factor, so it got factored out.
It was the greatest linear factor. It is not a monomial, it is a binomial, just like [itex]x^2+ 9[/itex].

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 [tex]x^2 \bold{(x + 5)} + 9 \bold{(x + 5)} = 0[/tex] and you get [tex](x^2 + 9)(x + 5) = 0[/tex]. Maybe part of the confusion is that the GCF appears last instead of first?


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