0

My teacher found this cool shortcut for factoring. I would like to use, for it saves time, but I feel hesitant using it without knowing the mathematical proof. Can anyone watch the video and explain it to me, thanks. The link to the video is here, http://m.youtube.com/watch?v=r1JAJfmRG5w.

Sam
  • 693
  • The answer by Gone in this question may be what you are looking for: http://math.stackexchange.com/questions/138925/factoring-ac-to-factor-ax2bxc?rq=1 – tylerc0816 Nov 21 '13 at 03:44

1 Answers1

1

Start with any quadratic with integer coefficients: $$Ax^2+Bx+C$$

Now let's follow what your teacher did.

Step 1. Write $x^2+Bx+AC$.

Step 2. Factor $x^2+Bx+AC=(x-q)(x-p)$.

Step 3. Write $(x-\frac{q}{A})(x-\frac{p}{A})$ and reduce the fractions $\frac{q}{A}=\frac{r}{s}$ and $\frac{p}{A}=\frac{t}{k}$.

Step 4. The desired factorizations is $(sx-r)(kx-t)$.

To prove this works, note from step 2 that $B=-(q+p)$ and $AC = pq$. This is sufficient information to obtain the roots of $Ax^2+Bx+C$ by the quadratic formula: $$ x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}= \frac{q+p\pm \sqrt{(p+q)^2-4pq}}{2A}= \frac{q+p\pm \sqrt{(q-p)^2}}{2A}.$$ So the roots are $\frac{q}{A}$ and $\frac{p}{A}$. This agrees with what we found above, since $\frac{q}{A}=\frac{r}{s}$ and $\frac{p}{A}=\frac{t}{k}$.

Doc
  • 1,966