Functions

**EmG** · December 14th 2009, 01:47 AM

Anyone know this one?

$f(x)=9x^3-48x^2+64x$

**Raoh** · December 14th 2009, 01:54 AM

Originally Posted by EmG

Anyone know this one?

$f(x)=9x^3-48x^2+64x$

hello
factorize with x.
$x(9x^{2}-48x+64)=x(8-3x)^{2}$

**EmG** · December 14th 2009, 01:55 AM

I did and when I did it I get stuck at $f(x)=x(9x^2-48x+64)$

**Raoh** · December 14th 2009, 01:58 AM

Originally Posted by EmG

I did and when I did it I get stuck at $f(x)=x(9x^2-48x+64)$

$9x^2-48x+64=(3x)^2-48x+8^2$
is it clear now ?

**Raoh** · December 14th 2009, 02:00 AM

Remember !
$(a-b)^2=a^2-2ab+b^2$

**EmG** · December 14th 2009, 02:02 AM

Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.

**Raoh** · December 14th 2009, 02:08 AM

Originally Posted by EmG

Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.

okay..
do you agree that $f(x)=x(8-3x)^{2}$
if yes,all you have to do is to put f(x)=0 and find the solutions.

**EmG** · December 14th 2009, 02:16 AM

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?

**Raoh** · December 14th 2009, 02:22 AM

Originally Posted by EmG

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?

$f(x)=0\Leftrightarrow x(8-3x^2)=0$
therefore the solutions are $\left [ x=0,x=\frac{8}{3} \right ]$ .
hope that helps.

**mr fantastic** · December 14th 2009, 03:10 AM

Originally Posted by EmG

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?

Go back and review the basic theory that is undoubtedly given in your classnotes or textbook. You're expected to know that

$f(x)=0 \Rightarrow x(8-3x^2)=0 \Rightarrow x = 0 \, \text{or} \, 8 - 3x^2 = 0 \Rightarrow x = ....$

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