How do you factor #20x^2 + 37x + 15 #?

3 Answers
Aug 5, 2015

Answer:

Use the quadratic formula to find the roots of #20x^2+37x+15=0# and hence that:

#20x^2+37x+15 = (4x+5)(5x+3)#

Explanation:

Let #f(x) = 20x^2+37x+15#

This is of the form #ax^2+bx+c#, with #a=20#, #b=37# and #c=15#.

This has determinant #Delta# given by the formula:

#Delta = b^2-4ac = 37^2 - (4xx20xx15) = 1369-1200 = 169 = 13^2#

Since this is a perfect square, #f(x) = 0# has rational roots and #f(x)# has factors with rational coefficients.

Since we have gone to the trouble of computing the determinant, we might as well use the quadratic formula to find the roots of #f(x) = 0# as:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a) = (-37+-13)/40#

That is: #x = (-37-13)/40 = -50/40 = -5/4#

or: #x = (-37+13)/40 = -24/40 = -3/5#

Hence #f(x) = (4x+5)(5x+3)#

Aug 5, 2015

Answer:

Use a version of the AC Method to find:

#20x^2+37x+15=(4x+5)(5x+3)#

Explanation:

Let #A=20#, #B=37#, #C=15#.

Look for a factorization of #AC=20*15=300=2^2*3*5^2# into a pair of factors #B1# and #B2# whose sum is #B=37#.

Note that one of #B1# and #B2# must be divisible by #25# since otherwise both #B1# and #B2# would be divisible by #5# and so would their sum, but #37# is not divisible by #5#.

Try #B1=25#. Then #B2=300/(B1)=300/25=12#

That works: #25+12 = 37#.

Next, for each of the pairs #(A, B1)# and #(A, B2)# divide by the HCF (highest common factor) to get a pair of coefficients of a factor of our original quadratic...

#(A, B1) = (20, 25) -> (4, 5) -> (4x+5)#

#(A, B2) = (20, 12) -> (5, 3) -> (5x+3)#

So #20x^2+37x+15=(4x+5)(5x+3)#

Aug 5, 2015

Answer:

Factor y = 20x^2 + 37x + 15

Ans: (5x + 3)(4x + 5)

Explanation:

#y = 20x^2 + 37x + 15 =# 20(x - p)(x - q)

I use the new AC Method to factor trinomials.
Converted trinomial #y' = x^2 + 37 x + 300 =# (x - p')(x - q').
Factor pairs of (300) -> ...(6, 50)(10, 30)(12, 25). This sum is 35 = b.
Then, p' = 12, and q' = 25.
Therefor, #p = (p')/a = 12/20 = 3/5,# and #q = (q')/a = 25/20 = 5/4.#
Factored form: #y = 20(x + 3/5)(x + 5/4) = (5x + 3)(4x + 5)#