How do you factor $8 x^{2} + 6 x + 1$ $8x^2+6x+1$ ?

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Answer 1 · 2015-09-23T12:49:42

$= (2 x + 1) (4 x + 1)$ $=color(blue)((2x+1) (4x +1)$

$8 x^{2} + 6 x + 1$ $8x^2 + 6x +1$

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:

$N_{1} \cdot N_{2} = a \cdot c = 8 \cdot 1 = 8$ $N_1*N_2 = a*c = 8*1 =8$
and
$N_{1} + N_{2} = b = 6$ $N_1 +N_2 = b = 6$

After trying out a few numbers we get $N_{1} = 2$ $N_1 = 2$ and $N_{2} = 4$ $N_2 =4$

$2 \cdot 4 = 8$ $2 *4 = 8$ and $2 + 4 = 6$ $2 + 4= 6$

$8 x^{2} + 6 x + 1 = 8 x^{2} + 2 x + 4 x + 1$ $8x^2 + color(blue)(6x) +1 =8x^2 + color(blue)(2x +4x) +1$

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

$= (2 x + 1) (4 x + 1)$ $=color(blue)((2x+1) (4x +1)$

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