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 $1$.

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step-by-step explanation: