The expression $x^2 + 3x - 28$ can be written as $ (x + a) (x - b),$ and the expression $x^2 - 10x - 56$ written as $ (x + 2b) (x + c) $, where $a$, $b$, and $c$ are integers such that $c > 0.$ What is the value of $2c - a$?

Therefore the value of $2c-a$ is $1$.

Step-by-step explanation:

Given expression, $x²+3x-28$

=$x²+7x-4x-28$ [28=7*2*2, 7 - (2*2) = 7-4=3]

=$x (x+7) - 4 (x+7) $

=$ (x+7) (x-4) $

Given that $x²+3x-28$ can be written as $ (x+a) (x-b) $.

So comparing $ (x+7) (x-4) $ and $ (x+a) (x-b) $.

Therefore $a$ = $7$ and $b$ = 4

Again,$ x²-10x-56$

=$x²-14x+4x-56$ [ 56 = 7*2*2*2, - (7*2) + (2*2) = - 14+4=10]

=$x (x-14) + 4 (x-14) $

=$ (x-14) (x+4) $

Given that $x²-10x-56$ can be written as $ (x+2b) (x+c) $.

So comparing $ (x-14) (x+4) $and$ (x+2b) (x+c) $.

Then $c$=$4$ [∵$c>0$]

Therefore $2c-a$

= $ (2*4) - 7$

=$8-7$

=$1$

Therefore the value of $2c-a$ is $1$.