How do you factor the trinomial #20x² + 47x − 24#?

1 Answer
George C.
Feb 8, 2017

#20x^2+47x-24 = 20(x+47/40-sqrt(4129)/40)(x+47/40+sqrt(4129)/40)#

Explanation:

Use the quadratic formula to find the zeros and hence the linear factors...

#20x^2+47x-24#

is in the form:

#ax^2+bx+c#

with #a=20#, #b=47#, #c=-24#

Let us take a quick look at the discriminant to decide how we should solve this:

#Delta = b^2-4ac = 47^2-4(20)(-24) = 2209+1920 = 4129#

Since #Delta > 0# this quadratic has two distinct Real zeros and will factor as a product of two linear terms with Real coefficients.

Note that #4129# is not a perfect square - it is a prime number - so the zeros are irrational. That means that we will not find any factors using an AC method or similar. We could use completing the square, but the coefficients #20# and #47# make that a little painful.

So let us use the quadratic formula...

The zeros given are by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (-47+-sqrt(4129))/40#

#color(white)(x) = -47/40+-sqrt(4129)/40#

Hence we can factor the given quadratic as:

#20x^2+47x-24 = 20(x+47/40-sqrt(4129)/40)(x+47/40+sqrt(4129)/40)#

Related questions
See all questions in Factorization of Quadratic Expressions
Impact of this question
1214 views around the world
You can reuse this answer
Creative Commons License