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I have a test tomorrow dealing with the factoring of trinomials, difference of squares, and sum/difference of cubes.

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- Thread starter wScott
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- #1

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I have a test tomorrow dealing with the factoring of trinomials, difference of squares, and sum/difference of cubes.

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mathwonk

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but a polynomila has a factor x-a of degree one if and only if it has a root equal to a. now rational roots of integer polynomilas must be of form c/d where c is a factor of the constant term and d is a factor of the elading coefficient.

so that helps find all factors of integer polynomilas with rational coefficients.

so learn the "root / factor" theorem.l

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With cubes we have a^3-b^3=(a-b)(a^2+ab+b^2), and a^3+b^3= (a+b)(a^2-ab+b^2).

For trinominals, we look at things like (x-1)(x-6)=X^2-7x+6. You can use the root/factor theorem here. Simply solve for the roots of X^2-7x+6 = 0.

If its rational, it must be an integer since the leading term is unity, so we could try plus or minus: 1,2,3,6. You need practice on these kind of problems.

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I did good on the test but could you guys tell me what the root / factor theorem is? It sounds interesting.

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Mathwonk described the theorem in his post. For polynomials of the second degree, the result is trivial, so it is generally called upon to reduce higher degree polynomials with rational roots. Plug in rational combinations of factors of the constant term over factors of the leading coefficient until you get a zero, then use that root in a binomial factor to reduce the polynomial by one degree. Reduction is effected with long division of the larger polynomial by the binomial factor, if any rational factor exists.

For example, you want to find the zeros of the equation 2x^3 + 3x^2 - 4x - 1 = 0. By the theorem, if this equation has rational zeros, they can only be of the forms 1/2 and 1/1=1. Trying each term out, we find x=1 is a zero, so (x-1) is a factor of the polynomial. Long division gives 2x^3 + 3x^2 - 4x - 1 = (x-1)(2x^2 + 5x + 1). The quadratic is then easily factored to give the remaining irrational zeros of the original equation.

For example, you want to find the zeros of the equation 2x^3 + 3x^2 - 4x - 1 = 0. By the theorem, if this equation has rational zeros, they can only be of the forms 1/2 and 1/1=1. Trying each term out, we find x=1 is a zero, so (x-1) is a factor of the polynomial. Long division gives 2x^3 + 3x^2 - 4x - 1 = (x-1)(2x^2 + 5x + 1). The quadratic is then easily factored to give the remaining irrational zeros of the original equation.

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mathwonk

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maybe by another name? it is a basic sophomore high school theorem that says a is a root iff x-a is a factor, ring a bell?

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Nope, sorry mathwonk but I don't know that.

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mathwonk

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other direction: divide f(x) by x-a and get f(x) = (x-a)q(x) + r(x) where r is the remiander, and hence must have degree less than x-a, so muist have degree zero, i.e. r is a constant. then set x =a and get that r = f(a).

so x-a divides f if and only if the remainder on division by x-a equals zero, but since that remainder is f(a), x-a divides f(x) if and only if f(a) = 0, i.e. if and only if a is a root.

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Tom Mattson

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