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For an assignment in my Algebra 2 class we have to find the rational zeros of the equation then write it in factored form.

I am a little lost and don't really get it

Equation:

$h(x)=x^3-5x^2+2x+8$

I have tried first getting the rational zeros and got

$1, 2, 4, 8$

Then using synthetic substition

$2$ | $1$ $-5$ $2$ $8$

____2 -6 -8____

From that i got

$1x^3$ $-3$ $-4$ and finally $0$

Then wrote it into a function

$h(x)=(x-2)(x^2-3x-4x)$

Which i don't know if it is correct, i also don't know what they mean by factor it

It would be helpful if you could show all you steps so i can actaully learn how to do it myself.

Thanks in advance!

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    $\begingroup$ You have factorised it correctly. We have $h (2)=0$ so one of the roots is 2. Also the quadratic should be $x^2-3x-4$. $\endgroup$ – user371838 Dec 8 '16 at 2:53
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    $\begingroup$ One mistake: it's $x^2-3x-4$, not $x^2-3x-4x = x^2-7x$. $\endgroup$ – 2012ssohn Dec 8 '16 at 2:54
  • $\begingroup$ A polynomial of degree 3 with 4 roots ? Is that what you mean when getting the zeros ? $\endgroup$ – Aseed Dec 8 '16 at 3:05
  • $\begingroup$ @Xxdzs sorry meant rational zeros $\endgroup$ – Kingbluesapphire Dec 8 '16 at 3:06
  • $\begingroup$ But by zero, you mean a value such that h(v) = 0 ?? Because h(1) = 1 - 5 + 2 + 8 = 6 != 0 Not sure to follow you $\endgroup$ – Aseed Dec 8 '16 at 3:07
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You're doing everything right so far! Just factor the quadratic $x^2-3x-4$ to completely factor $x^3-5x^2+2x+8$.


Although, you could do it this way: $$\begin{align*} & x^3-5x^2+2x+8\\ & =x^3-4x^2-x^2+2x+8\\ & =x^2(x-4)-(x-4)(x+2)\\ & =(x-4)(x^2-x-2)\\ & =(x-4)(x-2)(x+1)\end{align*}\tag{1}$$

But not many people might've seen the first step...

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