I'm interested to know some different methods to factor equations of the form $ax^2+bx+c$, where $a \ne 0$, other than pure guess and check.
$\begingroup$
$\endgroup$
13
-
1$\begingroup$ British Method is one way $\endgroup$– suomynonACommented Sep 22, 2016 at 23:13
-
$\begingroup$ What? Please explain $\endgroup$– thunderboltCommented Sep 22, 2016 at 23:13
-
$\begingroup$ @fleablood I only know how to use the binomial theorem to expand binomials $\endgroup$– thunderboltCommented Sep 22, 2016 at 23:18
-
$\begingroup$ Compute the roots $x_0$ and $x_1$ (poss.ibly equal) with the quadratic formula and factor as $a(x-x_0)(x-x_1)$. $\endgroup$– BernardCommented Sep 22, 2016 at 23:21
-
3$\begingroup$ @fleablood Didn't you say the binomial theorem earlier? What do you mean? $\endgroup$– thunderboltCommented Sep 22, 2016 at 23:23
|
Show 8 more comments
4 Answers
$\begingroup$
$\endgroup$
10
One method is The British Method. Here are the steps.
$1)$ Multiply $a*c$.
$2)$ Find a factor pair of $ac$ that adds up to $b$
$3)$ Replace $bx$ in the equation with the two factors, both multiplied by x.
$4)$ Factor by grouping.
Here is an annotated example. $$2x^2-x-6$$ $$a=2, b=-1, c=-6$$ Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$. The equation becomes $$2x^2-4x+3x-6$$ Then we factor by grouping $$2x^2-4x+3x-6$$ $$2x(x-2)+3(x-2)$$ $$(2x+3)(x-2)$$
Another method is shown below.
Steps:
$1)$ Multiply $a*c$.
$2)$ Find a factor pair of $ac$ that adds up to $b$, call them $x_1$ and $x_2$
$3)$ Write the following equation: $(ax+x_1)(ax+x_2)$.
$4)$ Factor out the GCF of each individual equation. The part remaining after factoring out the GCF is your answer.
Annotated example: $$2x^2-x-6$$ $$a=2, b=-1, c=-6$$ Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$, getting the equation: $$(2x-4)(2x+3)$$ Factor out a $2$ from the first part, where you get the final answer of $$(x-2)(2x+3)$$
answered Sep 22, 2016 at 23:20
-
2$\begingroup$ Yes, there is an element of guess and check involved, but OP wrote "pure guess and check" which I interpreted as methods that only used guess and check. $\endgroup$ Commented Sep 22, 2016 at 23:30
-
2$\begingroup$ If we aren't going to use some arithmetic guess and check step then quadratic formula or completing the square become your only options and for many questions those techniques are over the top. $\endgroup$ Commented Sep 22, 2016 at 23:32
-
1
-
4$\begingroup$ @fleablood Chill. I never said these methods were the best methods or the only methods. I'm just giving some interesting methods to answer OP's question. $\endgroup$ Commented Sep 23, 2016 at 0:30
-
1$\begingroup$ Thanks again, amazing answer =) $\endgroup$ Commented Sep 27, 2016 at 1:48
$\begingroup$
$\endgroup$
6
Assuming your question is in earnest and you aren't trolling the answer is
$ax^2 + bx + c = a(x + \frac{b + \sqrt{b^2 - 4ac}}{2a})(x + \frac{b - \sqrt{b^2 - 4ac}}{2a} )$
No guessing. If the equation factors at all (which it won't if $4ac > b^2$) then it will always factor to that.
answered Sep 22, 2016 at 23:20
-
3$\begingroup$ The OP isn't asking for different factorizations, but for different methods of arriving at a factorization. Certainly when you see $x^2 + 6x - 8$ you don't factor it using the quadratic formula, do you? $\endgroup$– mweissCommented Sep 22, 2016 at 23:23
-
1$\begingroup$ The op said "other than guessing and checking" I never guess. I solve and factor from that. If the OP knew how to solve there'd be no reason to ask the question. No, I personally find completely the square to be easier than the quadratic equation but .. why on earth would you be surprised if I did... Factoring is the same thing as solving. $\endgroup$ Commented Sep 22, 2016 at 23:27
-
2$\begingroup$ I apologize for presuming, but I think most people look at $x^2 + 6x - 8$ and think to themselves "What two numbers multiply to $-8$ and sum to $6$?" $\endgroup$– mweissCommented Sep 22, 2016 at 23:29
-
$\begingroup$ So ... given $x^2 + 6x - 8$ I'd divide the 6 to get 3 and square it to get 9 subtract 9 from -8 to to get -17. Switch signs and take the two square roots and and three to get $(x - 3+ \sqrt(17))(x - 3 - \sqrt(17))$ ... which is the quadratic formula. $\endgroup$ Commented Sep 22, 2016 at 23:35
-
1$\begingroup$ No two rational numbers sum to -8 and sum to 6. If it wer $x^2 + 6x + 8$ I'd take a look at that but if before I go grubbing and guessing I want to know if there is a potato in that mud. $x^2 + 6x + 10$ won't have any and though $x^2 + 6 + 8$ will it's just as easy to square 3 and subtract 8 to get 1. That way I know it will factor. $x^2 + 6x + 10$ won't. $\endgroup$ Commented Sep 22, 2016 at 23:39
$\begingroup$
$\endgroup$
1
Variation of the method mentioned by @suomnonA
1) Multiply $a*c$.
2) Find a factor pair of $ac$ that adds up to $b$. Let them be $x_1$ and $x_2$.
3) Write factors as:
$$\frac{(ax-x_1)(ax-x_2)}{a}$$
The factors on top will factorize down to cancel out with the $a$ on the bottom.
Example:
$$6x^2+17x-14$$ $$a=6, b=17, c=-14$$ Here we do step $1$, getting $$ac=-84$$ Then we do step $2$. The factor pair is $$-4,21$$ Now we do step $3$.
The equation becomes $$\frac{(6x-4)(6x+21)}{2}$$
which factorizes to
$$\frac{2(3x-2)\cdot3(2x+7)}{6}$$
$$(3x-2)(2x+7)$$
answered Sep 22, 2016 at 23:28
-
$\begingroup$ Apparently this is referred to as the Airplane Method. $\endgroup$ Commented Sep 22, 2016 at 23:38
$\begingroup$
$\endgroup$
Another option:
$$a\,{{x}^{2}}+bx+c$$ $$\downarrow$$ $$n\,{{x}^{2}}-2knx+{{k}^{2}}n-m$$ $$\downarrow$$ $${{x}^{2}}-2kx-\frac{m}{n}+{{k}^{2}}$$ $$\downarrow$$ $$\left( x-\sqrt{\frac{m}{n}}-k\right) \,\left( x+\sqrt{\frac{m}{n}}-k\right) $$
