# Is it possible to factor y=x^2+x-12 ? If so, what are the factors?

Jul 7, 2016

$y = \left(x - 3\right) \left(x + 4\right) .$

#### Explanation:

Yes, $y = f \left(x\right) , s a y , = {x}^{2} + x - 12$ can be factorised into two linear factors.

In fact, a quadratic polynomial like $a {x}^{2} + b x + c$ can be factorised [ in RR], if and only if, $\Delta = {b}^{2} - 4 a c \ge 0.$

In our case, we are having, $a = 1 , b = 1 , c = - 12 ,$ so, $\Delta = 1 - 4 \cdot 1 \cdot \left(- 12\right) = 1 + 48 = 49 \ge 0$, hence, given $f$ can be factorised.

Its factors are $\left(x - \alpha\right) , \left(x - \beta\right)$, where, $\alpha = \frac{- b + \sqrt{\Delta}}{2 a} = \frac{- 1 + \sqrt{49}}{2 \cdot 1} = \frac{- 1 + 7}{2} = 3$

and, #beta=(-b-sqrtDelta0/(2a)=(-1-7)/2=-8/2=-4.

Hence the factors are $\left(x - 3\right) \mathmr{and} , \left(x + 4\right) .$

$\therefore y = \left(x - 3\right) \left(x + 4\right) .$

Another Method, is, as follows :-

$y = f \left(x\right) , s a y , = {x}^{2} + x - 12$
$\therefore y = {x}^{2} + 4 x - 3 x - 12. \ldots \ldots \ldots \ldots \ldots \ldots . . \left[4 \times 3 = 12 , 4 - 3 = 1\right]$
$= x \left(x + 4\right) - 3 \left(x + 4\right) = \left(x + 4\right) \left(x - 3\right)$, as before!

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Jul 7, 2016

$\left(x + 4\right) \left(x - 3\right)$

#### Explanation:

Find two factors of 12 which differ (because of the MINUS 12) by 1, as indicated by the coefficient of the x term.

With a basic knowledge of the times tables we find:

$4 \times 3 = 12 \text{ and } + 4 - 3 = + 1$
These then are the factors we need. The MINUS 12 also indicates that the sign will be different, (neg x pos = neg)

$\left(x + 4\right) \left(x - 3\right)$