anonymous
  • anonymous
How do you factor a perfect square trinominal?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
What's the trinomial?
anonymous
  • anonymous
x^2+bx+c and ax^2+bx+c
anonymous
  • anonymous
would like information ont he process so I can learn to do them myself correctly

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

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