How do you write $y=5x^2+5x-3$ in factored form?

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Answer 1 · 2015-09-22T13:11:25

You'll need to find the zeros first.

We cannot factor $y=5x^2+5x-3$ by trial and error, or by AC, or by grouping.

Solve: $5x^2+5x-3 = 0$ by either completing the square or by using the quadratic formula to get solutions

$x_1 = (-5+sqrt85)/10$ and $x_2 = (-5-sqrt85)/10$

The factors of $5x^2+5x-3$ are $x-x_1$ and $x-x_2$ , so we have the factored form:

$y = (x-(-5+sqrt85)/10)(x-(-5-sqrt85)/10)$

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If you wanted vertex form, here is the answer

$y = 5(x^2+x color(white)"xxxxxx")-3$

$y = 5(x^2+x +1/4)-3-5/4$

$y = 5(x-1/2)^2 - 17/4$

Answer 2 · 2015-09-22T13:21:51

Refer to explanation

First we have to find the roots of $5x^2+5x-3=0$ . Hence the roots are

$x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$

where $a=5,b=5,c=-3$ so we have that

$x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=> x_(1,2)=(-5+-sqrt(5^2+4*5*3))/10=(-5+-sqrt85)/10$

Hence the factor form is

$y=a*(x-x_1)*(x-x_2)=5*(x-(-5+sqrt85)/10)*(x-(-5-sqrt85)/10)= 1/20*(10x+5-sqrt85)*(10x+5+sqrt85)$

How do you write y=5x^2+5x-3y=5x^2+5x-3 in factored form?