Anyone know this one? $\displaystyle f(x)=9x^3-48x^2+64x$
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Originally Posted by EmG Anyone know this one? $\displaystyle f(x)=9x^3-48x^2+64x$ hello factorize with x. $\displaystyle x(9x^{2}-48x+64)=x(8-3x)^{2}$
I did and when I did it I get stuck at $\displaystyle f(x)=x(9x^2-48x+64)$
Originally Posted by EmG I did and when I did it I get stuck at $\displaystyle f(x)=x(9x^2-48x+64)$ $\displaystyle 9x^2-48x+64=(3x)^2-48x+8^2$ is it clear now ?
Remember ! $\displaystyle (a-b)^2=a^2-2ab+b^2$
Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.
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 $\displaystyle f(x)=x(8-3x)^{2}$ if yes,all you have to do is to put f(x)=0 and find the solutions.
When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?
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? $\displaystyle f(x)=0\Leftrightarrow x(8-3x^2)=0$ therefore the solutions are $\displaystyle \left [ x=0,x=\frac{8}{3} \right ]$. hope that helps.
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 $\displaystyle f(x)=0 \Rightarrow x(8-3x^2)=0 \Rightarrow x = 0 \, \text{or} \, 8 - 3x^2 = 0 \Rightarrow x = ....$
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