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My teacher taught me and my classmate to use cross-method to deal with quadratic polynomial, it is good if it involved smaller numbers like this: $$x^2+4x+3$$ after factorization, $$(x+3)(x+1)$$ If it involved a bigger number, it often wasted time to factorize it and I want to know if there is any faster method to factorize quadratic polynomial.

Max Chan
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    Well, the quadratic formula always works. – lulu Sep 29 '19 at 23:28
  • how? I know how to use it but how – Max Chan Sep 29 '19 at 23:28
  • If the lead coefficient is $1$ then $p(x)=(x-r_1)(x-r_2)$ where $r_1.r_2$ are the roots (as found by the quadratic formula, or however else). If the lead coefficient is not $1$ then factor it out. – lulu Sep 29 '19 at 23:32
  • I was just 15... Don't really understand – Max Chan Sep 29 '19 at 23:34
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    Ok. Consider your example, $p(x)=x^2+4x+3$. The quadratic formula says the two roots may be found as $\frac {-4\pm \sqrt {16-12}}2=\frac {-4\pm 2}2={-1,-3}$. Thus we should have $p(x)=(x-(-1))\times (x-(-3))=(x+1)(x+3)$ just as you wrote. This always works! – lulu Sep 29 '19 at 23:36
  • you can complete the square, or directly apply the quadratic formula, the quadratic formula will yield two solutions $x = r_1$ and $x=r_2$, with a little algebra you get $(x-r_1)(x-r_2)$ – Donlans Donlans Sep 29 '19 at 23:37
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    Aside from the three most commonly taught methods, there are a ton of methods to factorise quadratics. Therefore I recommend that people use this post instead so we can have all the methods in one place. – Toby Mak Sep 30 '19 at 03:35
  • In addition, if you are interested, here is a video containing some of the not-so-known ways to factor quadratics. – Toby Mak Sep 30 '19 at 03:41

2 Answers2

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There are three well known methods to factorize quadratic polynomials.

Method 1 (the FOIL method): The first and simplest method is known as FOILing (what you have labeled as the cross-method). To apply this method, you directly factor the quadratic into two binomials. You have done this by rewriting

$$x^2+4x+3=0$$ as $$(x+3)(x+1)=0$$ which will give two solutions for $x=-1$ and $x=-3$.

Method 2 (Complete the square): Another technique is known as completing the square. To do this suppose you are given a quadratic equation in the form

$$ax^2 + bx + c = 0 $$

then you can

  1. Divide all terms by $a$ (the coefficient of $x^2$).
  2. Move the number term ($c/a$) to the right side of the equation.
  3. Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
  4. Take the square root on both sides of the equation.
  5. Subtract the number that remains on the left side of the equation to find $x$.

in your case the quadratic equation is given as

$$x^2+4x+3=0$$

so if we complete the steps above then

  1. Since $a=1$, the quadratic is already in the desired form.
  2. We can rewrite the equation as $$x^2+4x=-3$$
  3. We can take $b/2=2$ and rewrite the LHS as $$(x+2)^2$$ and then balance this with the RHS by writing $$(x+2)^2=1$$
  4. Taking the square root of both sides forms $$x+2=\pm 1$$
  5. We can then subtract $2$ from both sides to form $$x=-2\pm 1$$ which gives us our two solutions of $x=-1$ and $x=-3$.

Method 3 (the quadratic formula): This method will solve any quadratic but is often more difficult than the first two methods. Note that this method may be necessary to find solutions to quadratics with imaginary roots. In general, if we are given a quadratic polynomial in the form

$$ax^2 + bx + c = 0$$

then the values of $x$ which are the solutions of the equation are given by

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

as we are given

$$x^2+4x+3=0$$

we see that $a=1,b=4,c=3$ so that

$$x=\frac{-4\pm\sqrt{16-4(1)(3)}}{2}=-2\pm\frac{\sqrt{4}}{2}=-2\pm1$$ which once again generates our two solutions of $x=-1$ and $x=-3$.

Axion004
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Lets use the quadratic formula to factor:

Take $(x^2+5x+6)$ You might be able to do this in your head but lets see how the quadratic formula would do it.

$\frac{-5+\sqrt{25-24}}{2}=-2$ $\frac{-5-\sqrt{25-24}}{2}=-3$. So $x=-3$ and $x=-2$ make this polynomial zero.

So it can be factored as $(x+2)(x+3)$. quadratics can be factored this way by qudratic formula . If you have $a$ not equal to zero, just factor it out and apply it at the end. Some roots may be non rational or complex.

Bernard
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Sina Babaei Zadeh
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