How do you factor the trinomial #6x^2-11x-35#?

2 Answers
Apr 9, 2016

# color(green)((3x + 5 ) ( 2x - 7 ) # is the factorised form of the expression.

Explanation:

#6x^2 - 11x - 35#

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

#N_1*N_2 = a*c = 6*(-35) = -210#

AND

#N_1 +N_2 = b = -11#

After trying out a few numbers we get #N_1 = 10# and #N_2 =-21#
#10* (-21) = -210#, and #10+(-21)= -11#

#6x^2 - 11x - 35 = 6x^2 - 21x + 10x - 35#

# = 3x ( 2x - 7 ) + 5 ( 2x - 7 )#

#(2x - 7 )# is a common factor to each of the terms

# = (3x + 5 ) ( 2x - 7 ) #

# color(green)((3x + 5 ) ( 2x - 7 ) # is the factorised form of the expression.

Apr 9, 2016

(3x + 5)(2x - 7)

Explanation:

Another factoring way is using the new AC Method (Socratic Search)
#y = 6x^2 - 11x - 35 =# 6(x + p)(x + q)
Converted trinomial #y' = x^2 - 11x - 210 =# (x + p')(x + q')
p' and q' have opposite signs because ac < 0.
Factor pairs of (-210) -->...(-10, 21)(10, -21). This sum is -21 = b. Then p' = 10 and q' = -21
Back to trinomial y -> p = (p')/a = 10/6 = 5/3 and q = (q')/a = -21/6 = -7/2
Factored form: y = 6(x + 5/3)(x - 7/2) = (3x + 5)(2x - 7)