How do you factorize quadratics when the coefficient of $x^2 \gt 1$?

Question

So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:

$2x^2-x-3$

Can anyone help me?

Answer 1

For quadratic polynomials of the form $f(x)=ax^2+bx+c$ with $a,b,c$ real numbers, the roots are known to be given by the following:

The Quadratic Formula: The roots of $f(x)=ax^2+bx+c$ are $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ and $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

Note that when $b^2-4ac=0$, then the root is "repeated" and when $b^2-4ac<0$ then the roots are non-real complex numbers.

The polynomial then can be factored as $ax^2+bx+c = a(x-x_1)(x-x_2)$

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For your specific example, $2x^2-x-3$, this is of the form $ax^2+bx+c$ with $a=2,~b=-1,~c=-3$.

Applying the formula above gives the two roots as:

$x_1=\frac{-(-1)+\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1+\sqrt{1+24}}{4}=\frac{1+5}{4}=\frac{3}{2}$

$x_2=\frac{-(-1)-\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1-\sqrt{1+24}}{4}=\frac{1-5}{4}=-1$

You have then that $2x^2-x-3=2(x-\frac{3}{2})(x-(-1))$, which can be rewritten as $=(2x-3)(x+1)$

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note: in some settings we are interested only in factorizations which involve only integers. It is possible in this setting then that no such factorizations exist and we say the polynomial is irreducible

Answer 2

You can either use the quadratic formula for a general quadratic $ax^2+bx+c$ which is $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ which I will say no more of as JMoravitz has rigorously explained it.

Or you can complete the square, or write $$2x^2 \color{blue}{-x}-3$$ $$=2x^2+\color{blue}{2x-3x}-3$$ $$=2x(x+1)-3(x+1)$$ $$=\color{red}{(2x-3)(x+1)}$$ Notice, in the part marked $\color{blue}{\mathrm{blue}}$ I have simply rewritten $-x$ as $2x-3x$.

From then on you simply take out common factors.

Note that this method is only valid when you have two numbers whose product is $-6$ and sum is $-1$.

Or, put in another way for the general quadratic equation $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$. Otherwise using the quadratic formula is the best method as it suits all scenarios.

For completing the square: $$2x^2 -x-3$$ $$=2\left(x^2 -\frac{x}{2}-\frac{3}{2}\right)$$ $$=2\left(\left(x -\frac{1}{4}\right)^2-\frac{1}{16}-\frac{3}{2}\right)$$ $$=2\left(\left(x -\frac{1}{4}\right)^2-\left(\frac{5}{4}\right)^2\right)\tag{1}$$ $$=2\left(\left(x -\frac{1}{4}+\frac{5}{4}\right)-\left(x -\frac{1}{4}-\frac{5}{4}\right)\right)\tag{2}$$ $$=2\left((x+1)\left(x -\frac{3}{2}\right)\right)$$ $$=\color{red}{(2x-3)(x+1)}$$ as before.

From $(1)$ to $(2)$ I used the difference of two squares formula such that $a^2-b^2=(a+b)(a-b) \space \space\forall a,b \in \mathbb{R}$

Hope this helps.

Answer 3

You can try factoring by inspection.

$$2x^2-x-3=(ax+b)(cx+d)$$

That's your general form, and we are going to guess that $a$, $b$, $c$ and $d$ are integers (otherwise factoring will be a real mess).

Considering FOIL,

$b\times d= -3$, so one of them is negative. If they are integers, they are $1$ and $-3$ or $-1$ and $3$.

$a\times c=2$, so they are $1$ and $2$.

Finally, $bc + ad = -1$

We can experiment with different combinations of $(ax+b)(cx+d)$ until we find the one that meets these criteria.

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

This method is is not so direct as it involves. Trial and error. But it may help you see what's going on when you factor, and may help you make sense of the more direct, but perhaps more abstract methods.

Answer 4

Depending on your level, you may not have seen the quadratic formula that JMoravitz and BLAZE posted. That formula is often the way to find the zeroes of and factor a quadratic.

But what Adam suggests may be more up your alley, and it's what I'll discuss as well: You can make educated guesses and then check to see if it works. The problem you have posted does not have many options to guess, so this method is fairly quick.

Consider your quadratic and what you expect the factored form to look like:

$$2x^2-x-3 = (\color{red}{\Box}x \color{red}{\pm} \color{red}{\Box})(\color{red}{\Box}x\color{red}{\pm}\color{red}{\Box})$$

You know that the factored form requires you to fill in those blanks, choosing the appropriate sign and number. What numbers can possibly satisfy the equation?

Start with the $2x^2$ term on the left. This one is straightforward. In order to get $2x^2$, the coefficients of the $x$'s on the right hand side must be $2$ and $1$ if we are to have all integer coefficients.

$$2x^2-x-3 = (\color{green}{2}x \color{red}{\pm} \color{red}{\Box})(\color{green}{1}x\color{red}{\pm}\color{red}{\Box})$$

$$2x^2-x-3 = (2x \color{red}{\pm} \color{red}{\Box})(x\color{red}{\pm}\color{red}{\Box})$$

Now what about the $-3$ on the right? It's factors are $\pm1, \pm3$. We need to select two of these numbers and have them fill the remaining sections. You can try all combinations until you have the answer, but we can be smarter than that.

Since $-3$ is negative, one of these numbers will be positive and the other will be negative. Furthermore, the two factors will need to add to $-1$, after one of them is multiplied by $2$, since we have $-x$ on the left.

Notice that $2-3 = -1$. That's what we want. So the $2x$ multiplies a $1$ and the $x$ multiplies a $-3$.

$$2x^2-x-3 = (2x \color{green}{-3})(x\color{green}{+1})$$

We do a quick check:

$$(2x-3)(x+1) = 2x^2 +2x -3x -3 = 2x^2 -x -3$$

It works.

Answer 5

We can factor $2x^2 - x - 3$ with respect to the rationals if we can find two numbers with product $2 \cdot -3 = -6$ and sum $-1$. The factors of $-6$ are \begin{align*} -6 & = 1 \cdot -6 & -6 & = -1 \cdot 6\\ & = \color{blue}{2 \cdot -3} & & = -2 \cdot 3 \end{align*} Of these four pairs of factors, only $2$ and $-3$ have sum $-1$. Hence, \begin{align*} 2x^2 - x - 3 & = 2x^2 + 2x - 3x - 3 && \text{split the linear term}\\ & = 2x(x + 1) - 3(x + 1) && \text{factor by grouping}\\ & = (2x - 3)(x + 1) && \text{extract the common factor} \end{align*} Note that if we multiply $2x - 3$ and $x + 1$, we carry out the steps of the factorization in reverse order.

Why does this work? Suppose we have the factorization with respect to the rationals \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su} \end{align*} Observe that the product of the coefficients of the quadratic and constant terms is equal to the product of the two coefficients that sum to the coefficient of the linear term, that is
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st}) = rstu$$ Matching coefficients gives $a = rt$, $b = ru + st$, and $c = su$. Thus, we can factor a quadratic with respect to the rationals if we can find two numbers with product $ac$ and sum $b$.

Answer 6

You can get a monic polynomial out of his by putting $y=2x$. The expression becomes $$\frac12(y^2-y-6)$$ which you can easily factorise as $$\frac12(y-3)(y+2)$$ Substituting back gives $$(2x-3)(x+1)$$