## anonymous 5 years ago How do you factor a perfect square trinominal?

1. anonymous

What's the trinomial?

2. anonymous

x^2+bx+c and ax^2+bx+c

3. anonymous

would like information ont he process so I can learn to do them myself correctly

4. anonymous

Well, I think you're asking for information on how to factor trinomials, rather than perfect square trinomials.

5. anonymous

Ok yes sorry

6. anonymous

Really, all we need to discuss is the first one, since you can obtain the first from the second by factoring out the coefficient of x^2.

7. anonymous

Like the binominal process

8. anonymous

We'll see...

9. anonymous

Consider $x^2+bx+c$

10. anonymous

If this equation is to be factored, it will be in the form,$(x-\alpha)(a- \beta)$where alpha and beta are the roots (this is by the unique factorisation theorem). Expanding the product out, you'll have$x^2-(\alpha + \beta)x+ \alpha \beta$

11. anonymous

This is identical to the original expression, x^2 + bx + c, if and only if the coefficients according to each power are the same; that is,$\alpha + \beta =-b$and$\alpha \beta = c$

12. anonymous

These equations relate the roots to the coefficients.

13. anonymous

So, from the first,$\beta = -b - \alpha$and substituting into the second,$c=\alpha \beta = \alpha (-b-\alpha)=-b \alpha-\alpha^2 \rightarrow \alpha^2+b \alpha + c =0$

14. anonymous

So to find alpha, we need to solve this quadratic equation. You can do this by completing the square, or using the quadratic formula. They'll both lead you to,$\alpha = \frac{-b \pm \sqrt{b^2-4c}}{2}$If you take the first solution,$\alpha = \frac{-b + \sqrt{b^2-4c}}{2}$and plug it into any one of the expressions linking the roots with the coefficients, you'll find,$\beta = \frac{-b - \sqrt{b^2-4c}}{2}$

15. anonymous

Similarly, if you had chosen the negative solution for alpha (i.e. - between b and square root), beta would have been the positive. It therefore makes no difference which you choose to start with.

16. anonymous

So, then, the factorisation of your quadratic is$x^2+bx+c$ $=\left( x-\left( \frac{b+\sqrt{b^2-4c}}{2} \right) \right)\left( x-\left( \frac{b-\sqrt{b^2-4c}}{2} \right) \right)$

17. anonymous

what happens if the coefficients according to each power are different, is there more than one way to factor?

18. anonymous

This will give you the answer no matter what your coefficients are. If, in your second case, you have ax^2, you can see,

19. anonymous

$ax^2+bx+c=a \left( x^2 + \frac{b}{a}x +\frac{c}{a}\right)$

20. anonymous

The 'b' in the factorisation above is identified with b/a, and the c with c/a (i.e. replace b in the factored form with b/a and c with c/a).

21. anonymous

This will give you the factors no matter what coefficients you have.

22. anonymous

If you have special cases, then there may be simpler methods available to you, like 'cross method' and 'sum and product', but they're not that easy to show with algebraic expressions; they're more suited to numerical example.

23. anonymous

Ok... Thank You! :)

24. anonymous

You're welcome :) It's good you wanted to know!

25. anonymous

By the way, the 'cross method' and 'sum and product' may be found on YouTube or khanacademy.org.

26. anonymous

Ok great! I will check those out