How do you factor a perfect square trinominal?

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How do you factor a perfect square trinominal?

Mathematics
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What's the trinomial?
x^2+bx+c and ax^2+bx+c
would like information ont he process so I can learn to do them myself correctly

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Well, I think you're asking for information on how to factor trinomials, rather than perfect square trinomials.
Ok yes sorry
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.
Like the binominal process
We'll see...
Consider \[x^2+bx+c\]
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\[x^2-(\alpha + \beta)x+ \alpha \beta\]
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\[\alpha \beta = c\]
These equations relate the roots to the coefficients.
So, from the first,\[\beta = -b - \alpha\]and substituting into the second,\[c=\alpha \beta = \alpha (-b-\alpha)=-b \alpha-\alpha^2 \rightarrow \alpha^2+b \alpha + c =0\]
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,\[\beta = \frac{-b - \sqrt{b^2-4c}}{2}\]
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.
So, then, the factorisation of your quadratic is\[x^2+bx+c\] \[=\left( x-\left( \frac{b+\sqrt{b^2-4c}}{2} \right) \right)\left( x-\left( \frac{b-\sqrt{b^2-4c}}{2} \right) \right)\]
what happens if the coefficients according to each power are different, is there more than one way to factor?
This will give you the answer no matter what your coefficients are. If, in your second case, you have ax^2, you can see,
\[ax^2+bx+c=a \left( x^2 + \frac{b}{a}x +\frac{c}{a}\right)\]
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).
This will give you the factors no matter what coefficients you have.
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.
Ok... Thank You! :)
You're welcome :) It's good you wanted to know!
By the way, the 'cross method' and 'sum and product' may be found on YouTube or khanacademy.org.
Ok great! I will check those out

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