How do you make "-5(t^2)+10t+35" into factored form?

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- Apr 12th 2010, 01:56 PMsinjid9Factored form question
How do you make "-5(t^2)+10t+35" into factored form?

- Apr 12th 2010, 02:26 PMpickslides

First you take out a common factor of $\displaystyle -5$ leaving

$\displaystyle -5t^2+10t+35 = -5(t^2-2t-7)$

Now I would complete the square inside the brackets.

$\displaystyle -5(t^2-2t-7)$

You take the coeffeicent of the linear term, half it then square it.

$\displaystyle \left( \frac{-2}{2}\right)^2 = (-1)^2 = 1$ giving

$\displaystyle -5((t^2-2t{\color{red}+1})-7{\color{red}-1})$

$\displaystyle -5((t-1)^2-8)$

$\displaystyle -5((t-1)^2-(\sqrt{8})^2)$

Now applying the difference of 2 squares and we are finished.

$\displaystyle -5(t-1-\sqrt{8})(t-1+\sqrt{8})$ - Apr 12th 2010, 02:38 PMsinjid9
in the part:

http://www.mathhelpforum.com/math-he...3c07e363-1.gif

how do you know where to put the brackets that seperate (+1) and (-7) - Apr 12th 2010, 02:58 PMpickslides
In $\displaystyle -5(t^2-2t-7)$ I was looking for a new constant term that would help me make a perfect square. Finidng $\displaystyle +1$ as this term gave

$\displaystyle -5((\underbrace{t^2-2t{\color{red}+1}}_{\text{perfect square}})-7{\color{red}-1})$

I couldn't just throw away the $\displaystyle -7$ and to balance the equation I introduced an extra $\displaystyle -1$ from the $\displaystyle +1$ found.

A perfect square has the from of either

$\displaystyle a^2-2ab+b^2 = (a-b)^2$ or $\displaystyle a^2+2ab+b^2 = (a+b)^2$ - Apr 12th 2010, 03:33 PMsinjid9
k one more question. If I were to make this into an equation like this

"h=-5(t^2)+10t+35" is it possible to make "h=-5(t^2)+10t+35" into a(x-s)(x-t)? Without graphing it? - Apr 12th 2010, 03:56 PMpickslides