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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's the trinomial?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2+bx+c and ax^2+bx+c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would like information ont he process so I can learn to do them myself correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I think you're asking for information on how to factor trinomials, rather than perfect square trinomials.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Like the binominal process

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Consider \[x^2+bx+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0This 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
 5 years ago
Best ResponseYou've already chosen the best response.0These equations relate the roots to the coefficients.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So 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,\[\alpha = \frac{b \pm \sqrt{b^24c}}{2}\]If you take the first solution,\[\alpha = \frac{b + \sqrt{b^24c}}{2}\]and plug it into any one of the expressions linking the roots with the coefficients, you'll find,\[\beta = \frac{b  \sqrt{b^24c}}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Similarly, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So, then, the factorisation of your quadratic is\[x^2+bx+c\] \[=\left( x\left( \frac{b+\sqrt{b^24c}}{2} \right) \right)\left( x\left( \frac{b\sqrt{b^24c}}{2} \right) \right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what happens if the coefficients according to each power are different, is there more than one way to factor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This will give you the answer no matter what your coefficients are. If, in your second case, you have ax^2, you can see,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ax^2+bx+c=a \left( x^2 + \frac{b}{a}x +\frac{c}{a}\right)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The '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
 5 years ago
Best ResponseYou've already chosen the best response.0This will give you the factors no matter what coefficients you have.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome :) It's good you wanted to know!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way, the 'cross method' and 'sum and product' may be found on YouTube or khanacademy.org.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok great! I will check those out
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