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

2 Answers
Sep 22, 2015

Answer:

You'll need to find the zeros first.

Explanation:

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:

Refer to explanation

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)#