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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?

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  • $\begingroup$ I recommend writing the factors of the coefficient of $x^2$ and the factors of the constant and then trying combinations of them $\endgroup$
    – tzamboiv
    Oct 13 '15 at 0:11
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    $\begingroup$ Note that $2x^2-x-3 = 2(x^2-\frac{1}{2}x-\frac{3}{2})$. If you say you know how to factor monic quadratics, then you should be able to factor what is in the parenthesis above. $\endgroup$
    – JMoravitz
    Oct 13 '15 at 0:21
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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 $ax^2+bx+c = a(x-x_1)(x-x_2)$


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


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

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  • $\begingroup$ Using that formula made no sense at all to me. I get $b^2 - 4ac$ to be 23, and the square root of that is just an absurdly complex number. I don't understand how I'm supposed to factor anything with this? $\endgroup$ Oct 13 '15 at 0:45
  • $\begingroup$ @Julian $b=-1,~a=2,~c=-3$. You get then $b^2-4ac = (-1)^2-4(2)(-3) = 1+24=25$, not $23$. $\endgroup$
    – JMoravitz
    Oct 13 '15 at 0:47
  • $\begingroup$ Also, since the term complex has mathematical meaning other than what you are using it for here, it is recommended to call it "complicated" as opposed to complex. $\endgroup$
    – JMoravitz
    Oct 13 '15 at 0:48
  • $\begingroup$ I failed to realise that $b^2$ gives me $(-1)^2$ and not $-1^2$. I see now that it is indeed 25. Meaning x1 = $\frac{6}{4}$ and x2 = $1$? $\endgroup$ Oct 13 '15 at 0:56
  • $\begingroup$ Indeed. Although many methods exist, some of the cleaner solutions like BLAZE's above require some intuition. I prefer teaching this one since it is very algorithmic and requires only routine application of the rule. In reality, this is taking the "completing the square" technique and coming up with a generic formula to follow. Be warned however that this only works for quadratics. You cannot use this formula to factor something like $ax^3+bx+c$. There do exist longer, more complicated formulae for cubics and quartics, but hardly anyone memorizes them. $\endgroup$
    – JMoravitz
    Oct 13 '15 at 1:00
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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.

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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.

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  • $\begingroup$ Are there like a million different methods of doing this? I am yet to find a single coherent one. I don't quite understand what you did here. Is this the favourable method? The quadratic formula as suggested by JMoravitz here got me nowhere. :( $\endgroup$ Oct 13 '15 at 0:47
  • $\begingroup$ @JulianNikolayKrogh-Fredrikse Yes there are several [maybe not a million :)] ways of doing this as you can also complete the square as well are you familiar with this? $\endgroup$
    – BLAZE
    Oct 13 '15 at 0:51
  • $\begingroup$ No I am not. Is that the "box method"? purplemath.com/modules/factquad2.htm Because that one was as incoherent to be as the ABC-formula.. I honestly didn't expect factoring at this level to be so hard.. $\endgroup$ Oct 13 '15 at 0:59
  • $\begingroup$ @JulianNikolayKrogh-Fredrikse If you tell me where you're confused I will explain further in my answer $\endgroup$
    – BLAZE
    Oct 13 '15 at 1:24
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    $\begingroup$ @Julian Note that, in BLAZE's first example, (s)he splits $-x = -1x$ up into $-1x = -3x + 2x$, where $-3 + 2 = -1,$ the coefficient of $x$. Note also that if you multiply $a = 2$ and $c = -3$, you get $2\cdot(-3) = -6$; that's how we know to use $-3$ and $2$ to split up $-x = -1x$. More detail here, this is known as factoring by grouping/"the AC method". It's equivalent to the box method, and relies on being able to factor polynomials with four terms by grouping. $\endgroup$
    – pjs36
    Oct 13 '15 at 1:52
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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.

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  • $\begingroup$ The quadratic formula is introduced pretty early, I thought. But this use of it may be something that OP hasn't seen. I've thought that maybe a diagram could be used to illustrate completing the square. $\endgroup$
    – MathAdam
    Oct 13 '15 at 3:37
  • $\begingroup$ @AdamHrankowski It is introduced pretty early. In the US, it's often shown in grade 7-9 depending on skill. That being said, there are classes at my university that have students who have yet to learn to factor. I hope my dialogue helps instead of hinders! There are a lot of ways to slice this cake :) $\endgroup$
    – suneater
    Oct 13 '15 at 3:43
  • $\begingroup$ I'm in BC, Canada, and I haven't seen it until Grade 11. $\endgroup$
    – MathAdam
    Oct 13 '15 at 3:45
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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 $$(2x-3)(x+1)$$

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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 $b$.

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