How do you factor the trinomial $4x^2-8x-5$ ?

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Answer 1 · 2016-03-15T15:24:57

$(2x -5)(2x + 1)$ is the factorised form of the expression.

$4x^2 -8x-5$

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:

$N_1*N_2 = a*c = 4*(-5) = -20$

AND

$N_1 +N_2 = b = -8$

After trying out a few numbers we get $N_1 = -10$ and $N_2 =2$

$2*(-10) = -20$ , and $2+(-10)= -8$

$4x^2 -8x-5 =4x^2 -10x+2x-5$

$= 2x(2x -5) +1 (2x -5)$

$(2x -5)$ is a common factor to each of the terms.

$(2x -5)(2x + 1)$ is the factorised form of the expression.

How do you factor the trinomial 4x^2-8x-54x^2-8x-5?