How do you factor the expressions #x^2+3x-4#?

2 Answers
Jun 22, 2018

#x_1=-4#
#x_2=1#,
You use the so called discriminant

Explanation:

Most of the quadratic equations have the form of: #ax^2+bx+c=0#
In your case: a=1 , b=3, c=-4
Then we use the discriminant #D=b^2-4*a*c#
In this case:
#D=3^2-4*1*(-4)#
We get the result: D=25
Now we need the roots, for which we use the equation:
#x_1=(-b-sqrt(D))/(2*a)#
#x_2=(-b+sqrt(D))/(2*a)#
We get:
#x_1=-4#
#x_2=1#
And that's all :)

Jun 22, 2018

#(x-1)(x+4)#

Explanation:

What tow numbers sum up to our middle term of #3#, and have a product of #-4# (the last term)?

After some trial and error, we arrive at #-1# and #4#. So we can factor our quadratic as

#(x-1)(x+4)#

Hope this helps!