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anonymous

  • 5 years ago

How do you factor a perfect square trinominal?

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  1. anonymous
    • 5 years ago
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    What's the trinomial?

  2. anonymous
    • 5 years ago
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    x^2+bx+c and ax^2+bx+c

  3. anonymous
    • 5 years ago
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    would like information ont he process so I can learn to do them myself correctly

  4. anonymous
    • 5 years ago
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    Well, I think you're asking for information on how to factor trinomials, rather than perfect square trinomials.

  5. anonymous
    • 5 years ago
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    Ok yes sorry

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Like the binominal process

  8. anonymous
    • 5 years ago
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    We'll see...

  9. anonymous
    • 5 years ago
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    Consider \[x^2+bx+c\]

  10. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    These equations relate the roots to the coefficients.

  13. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    what happens if the coefficients according to each power are different, is there more than one way to factor?

  18. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[ax^2+bx+c=a \left( x^2 + \frac{b}{a}x +\frac{c}{a}\right)\]

  20. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    This will give you the factors no matter what coefficients you have.

  22. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Ok... Thank You! :)

  24. anonymous
    • 5 years ago
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    You're welcome :) It's good you wanted to know!

  25. anonymous
    • 5 years ago
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    By the way, the 'cross method' and 'sum and product' may be found on YouTube or khanacademy.org.

  26. anonymous
    • 5 years ago
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    Ok great! I will check those out

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