Question

How do you factor a perfect square trinominal?

Answer 1

What's the trinomial?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Ok yes sorry

Answer 6

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.

Answer 7

Like the binominal process

Answer 8

We'll see...

Answer 9

Consider \[x^2+bx+c\]

Answer 10

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\[x^2-(\alpha + \beta)x+ \alpha \beta\]

Answer 11

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\[\alpha \beta = c\]

Answer 12

These equations relate the roots to the coefficients.

Answer 13

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

Answer 14

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,\[\beta = \frac{-b - \sqrt{b^2-4c}}{2}\]

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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).

Answer 21

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

Answer 22

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.

Answer 23

Ok... Thank You! :)

Answer 24

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

Answer 25

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

Answer 26

Ok great! I will check those out