How to factor $2 x^{2} + 3 x + 1 = 0$ $2x^2+3x+1=0$ ?

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Answer 1 · 2017-12-11T21:42:06

$(2 x + 1) (x + 1) = 0$ $(2x+1)(x+1) = 0$
$\Rightarrow x = - \frac{1}{2}, x = - 1$ $=> x = -1/2 , x= -1$

To solve this we must understand how to factor:

if $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$

Then you need to find two numbers who sum to make $b$ $b$ but multiply to make $a \cdot c$ $a*c$

So in this circumstance:

Plus to make $3$ $3$ and multiply to make $2 \cdot 1 = 2$ $2*1 = 2$

Pairs of numbers who multiply to make $2$ $2$ :

$(2, 1), (- 2, - 1)$ $( 2 , 1) , (-2 ,-1 )$

But out of these we see that $(2, 1)$ $(2,1)$ also plus to make $3$ $3$

So we can use the fact that $2 x + x = 3 x$ $2x + x = 3x$ :

$\Rightarrow 2 x^{2} + 2 x + x + 1$ $=> 2x^2 + 2x + x +1$

Now we can factor the first 2 and last 2 terms individually, but ensuring they both have the same factor:

$\Rightarrow 2 x (x + 1) + (x + 1)$ $=> 2x(x+1 ) + (x+1)$

Factoring out the $(x + 1)$ $(x+1)$ :

$\Rightarrow (x + 1) [2 x + 1]$ $=> (x+1)[2x + 1 ]$

$\Rightarrow (x + 1) (2 x + 1)$ $=> (x+1)(2x+1)$

If we want to solve for $0$ $0$ then either one of $(x + 1) or (2 x + 1)$ $(x+1) or (2x+1)$ must equal $0$ $0$ and $α \cdot 0 = 0$ $alpha*0 = 0$

So hence $(x + 1) = 0 \Rightarrow x = - 1$ $(x+1) = 0 => x = -1$

Also $(2 x + 1) = 0 \Rightarrow 2 x = - 1 \Rightarrow x = - \frac{1}{2}$ $(2x+1) = 0 => 2x = -1 => x = -1/2$

How to factor 2x^2+3x+1=02x2+3x+1=02x^2+3x+1=0 ?