Answer:
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 $1$.
