How do you factor $8 x^{2} + 14 x + 3$ $8x^2+14x+3$ ?

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Answer 1 · 2015-10-09T04:41:42

$(4 x + 1) (2 x + 3)$ $(4x+1)(2x+3)$

Split 14 in two parts such that sum is 14 and product is 24 as below:

$8 x^{2} + 12 x + 2 x + 3$ $8x^2 +12x +2x+3$

$(8 x^{2} + 12 x) + (2 x + 3)$ $(8x^2 +12x) + (2x+3)$

$4 x (2 x + 3) + (2 x + 3)$ $4x(2x+3) +(2x+3)$

$(4 x + 1) (2 x + 3)$ $(4x+1)(2x+3)$

Answer 2 · 2015-10-09T04:43:11

$(4 x + 1) (2 x + 3)$ $color(blue)( (4x+1)(2x+3)$ is the factorised form of the expression.

$8 x^{2} + 14 x + 3$ $8x^2+14x+3$

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

In this technique, if we have to factorise an expression like $a x^{2} + b x + c$ $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_{1} \cdot N_{2} = a \cdot c = 8 \cdot 3 = 24$ $N_1*N_2 = a*c = 8*3 = 24$

AND

$N_{1} + N_{2} = b = 14$ $N_1 +N_2 = b = 14$

After trying out a few numbers we get $N_{1} = 12$ $N_1 = 12$ and $N_{2} = 2$ $N_2 =2$
$12 \cdot 2 = 24$ $12*2 = 24$ , and $12 + 2 = 14$ $12+2= 14$

$8 x^{2} + 14 x + 3 = 8 x^{2} + 12 x + 2 x + 3$ $8x^2+color(blue)(14x)+3 =8x^2+color(blue)(12x+2x)+3$

$= 4 x (2 x + 3) + 1 (2 x + 3)$ $= 4x(2x+3) +1 (2x+3)$

$(2 x + 3)$ $(2x+3)$ is a common factor to each of the terms

$(4 x + 1) (2 x + 3)$ $color(blue)( (4x+1)(2x+3)$ is the factorised form of the expression.

How do you factor 8x2+14x+38x^2+14x+3?