Is it possible to factor #y=3x^2+11x-4 #? If so, what are the factors?

Question

Is it possible to factor #y=3x^2+11x-4   #? If so, what are the factors?

Avery Alexander · Accepted Answer

#y=(x+4)(3x-1)#

Use the AC method.

#y=3x^2+11x-4# is a quadratic equation in the form #ax+bx+c#, where #a=3, b=11, and c=-4#.

Multiply #a# times #c#.

#3xx-4=-12#

Find two numbers that when added equal #11# and when multiplied equal #-12#. The numbers #12# and #-1# meet the criteria.

Rewrite the equation substituting #12x# and #-x# for #11x#.

#3x^2+12x-x-4#

Group the terms into two groups of two terms.

#(3x^2+12x)-(x-4)#

Factor out #3x# from the first group and #-1# from the second group.

#3x(x+4)-(x+4)#

Factor out #(x+4)#.

#(x+4)(3x-1)#

#y=(x+4)(3x-1)#

Nathan Axel · Answer

#color(brown)((3x-1)(x+4))#

3 is prime so we can only have 1 and 3 as factors giving:

#(x+?)(3x+?)#

There are two sets of factors of 4 and they are: { 2,2} and {1,4}

Lets try the {1,4} combination and see what we get!

By the way, the constant of 4 in your question is negative so the #color(white)(....)#{1 ,4} have to be opposite in their sign.

We see that #3xx4=12# which is close to the #11# in #11x# so lets try that configuration:

#(x-4)(3x+1) = 3x^2-12x+x-4# Not quite correct as we have #-11x#.

Ok! Lets try something else. The 11 is the correct magnitude but the wrong sign. Lets try reversing the signs:

#(x+4)(3x-1) =3x^2+12x-x-4# Now we have it!

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So #color(white)(..)3x^2+11x-4 = color(brown)((x+4)(3x-1))#

Or if you like you can change the order of the brackets to.

#color(brown)((3x-1)(x+4))#

#color(blue)("They call this Commutative")#

in that they can commute/travel without changing the intrinsic value

Julia Adams · Answer

undefined Yes, it is possible to factor the quadratic equation $ y = 3x^2 + 11x - 4 $. The factors are $ (3x - 1)(x + 4) $.