How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
 A: You can use several methods to solve the equation. Firstly you can use the quadratic formula 
$$\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
or you can use decomposition. 
Let's try decomposition first. Look for two numbers that multiple to $-6$ and add to $-1$. These numbers would be $+2$ and $-3$. So, we would place those for $-x$ and have the following:
$$2x^2-3x+2x-3$$
Now here you can factor out $x$ from the first two terms.
$$x(2x-3)+(2x-3)$$
Notice the coefficient on $2x-3$ is $1$. So this means that our equation is $x+1$ and $2x-3$. Therefore, giving us the solution 
$$ (2x-3) = 0 \, \text{or} \, (x+1) = 0$$
Now, if you want to use the quadratic formula then you can do this. $a = 2$, $b = -1$ and $c = -3$. So we will have: 
$$\frac{6 \pm\sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$$
When you apply this formula twice, you will get the roots which are $x = \frac{3}{2}$ and $x = -1$. If you have any questions, let me know. I will try to clarify as much as possible.
A: find the roots(if any). this is the most general answer to these problems.
A: I have some more helpful and straight forward reasoning (Here is the step-by-step reasoning):
first we rewrite the term:
Suppose that A=$2x^2-x-3$ then
$$A=\frac{2(2x^2-x-3)}{2}$$
then simplify the last term to conclude a most-known identity:
$$\frac{2(2x^2-x-3)}{2}=\frac{(2x)^2-(2x)-6}{2}$$
let $y=2x;$ then we can rewrite again:
$$A=\frac{(y)^2-(y)-6}{2}$$
You can see that the numerator is a form of a most-known identity; then we have $y^2-y-6=(y+2)(y-3)$ and by rewriting again:
$$A=\frac{((2x)+2)((2x)-3)}{2}=\frac{2((x)+1)((2x)-3)}{2}=\frac{2(x+1)(2x-3)}{2}$$
finally we conclude the desired result:
$$A=(x+1)(2x-3)$$ 
A: Solution 1:
We usually do some tests by replacing x=0; x=1 and x=-1 in the equation (usual roots, you can take it as habit) here we can notice that -1 is a solution for the equation (root) so :
$$2x^2 - x - 3 =(x+1)\times 2 \times Q$$ and $$Q=ax+b;$$ 
Variables a and b can be extracted easily for this example (sometimes we can even do an euclidean division)
Result is :
$$ 2x^2 - x - 3 =(x+1)\times2\times(x-3)=2(x+1)(x-3)$$
Solution 2: 
The equation is 2nd degree one so either it has 1 double root, 2 roots or no solution in R
(1)  $$2x^2 - x - 3= ax^2 +bx+c=0$$  so a=2, b=-1 and c=-3
We calculate $$Delta=b^2-(4ac)=1-(4\times2\times(-3))=25 >0$$ 
We have two roots for equation 1 so we can write : $$2x^2 - x - 3 = a(x-x1)(x-x2)$$
$$x1=\frac{-b - \sqrt{Delta}}{2a} = \frac{(1-5)}{4}=-1;$$
$$x2=\frac{-b + \sqrt{Delta}}{2a} = \frac{(1+5)}{4}=3;$$
So 
$$2x^2 - x - 3 = a(x-x1)(x-x2) = 2(x+1)(x-3)$$
A: Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient. 
In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping:
\begin{align*}
2x^2 - x - 3 &= 2x^2 + (2-3)x - 3 \\ &= 2x^2 + 2x - 3x - 3 \\ &= 2x(x + 1) - 3(x + 1) \\ &= (2x - 3)(x+1).
\end{align*}

The rule comes from the distributive rule: $(ax + b)(cx + d) = (ac)x^2 + (ad + bc)x + bd$. The first coefficient is $ac$ and the last is $bd$, they multiply together to $abcd$, and the middle coefficient is $ad + bc$, the sum of a factor pair of $abcd$.

Another way to do it is to use the quadratic formula to find the roots of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so
$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm\sqrt{1 - -24}}{4} = \frac{1\pm 5}{4} = \frac{3}{2}, -1,$$
which gives a factorization into the monic polynomials $x+1$ and $2x - 3$.
A: Note that $2-3=-1,$ so that $$\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\ &= 2x^2+2x-3x-3\\ &= 2x(x+1)-3(x+1)\\ &= (x+1)(2x-3).\end{align}$$
