Intro to rational expressions
- Thread starter Illvoices
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Cubist
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Otis
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Hi Illvioces. If writing out a general method in words is sort of speaking, then, yes, Mark did that. He gave you the method.
If you're studying on your own, do you have a textbook? Here's a video that covers the method, with some examples.
In cases where the leading coefficient is not a=1, then we use a different method (factor by grouping).
?
HallsofIvy
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Note that, to do this, you pretty much have to assume that numerical factors are integers because you are basically using "trial and error" and if you allow non-integers there are simply too many to try! And most polynomials like this can't be factored with integer coefficients!
Of course, in order to factor, you need to know how to multiply! Others have mention "FOIL": (x+ a)(x+ b)= x^2+ ax+ bx+ ab= x^2+ (a+ b)x+ ab.
Your example is \(\displaystyle x^2- 2x- 8\). We need to find a and b such that ab= -8 and a+ b= -2. Possible factors are (1)(-8), (-1)(8), (2)(-4), and (-2)(4). The sums are, respectively, 1+ -8= -7, -1+ 8= 7, 2+ -4= -2, and -2+ 4= 2. We want -4+ 2 so our factors are (x- 4)(x+ 2)!
Check: \(\displaystyle (x- 4)(x+ 2)= x^2- 4x+ 2z- 8= x^2- 2x- 8\).
Written out in detail like that it is tedious but with practice you can do it quickly.
Of course, in order to factor, you need to know how to multiply! Others have mention "FOIL": (x+ a)(x+ b)= x^2+ ax+ bx+ ab= x^2+ (a+ b)x+ ab.
Your example is \(\displaystyle x^2- 2x- 8\). We need to find a and b such that ab= -8 and a+ b= -2. Possible factors are (1)(-8), (-1)(8), (2)(-4), and (-2)(4). The sums are, respectively, 1+ -8= -7, -1+ 8= 7, 2+ -4= -2, and -2+ 4= 2. We want -4+ 2 so our factors are (x- 4)(x+ 2)!
Check: \(\displaystyle (x- 4)(x+ 2)= x^2- 4x+ 2z- 8= x^2- 2x- 8\).
Written out in detail like that it is tedious but with practice you can do it quickly.
A mass of confusion.
First, what you have provided is a polynomial expression rather than a rational expression.
Second, any polynomial of degree n > 0 can be factored into n polynomials of degree 1, perhaps by using complex coefficients. Any polynomial of degree 2n > 0 and real coefficients can be factored into n polynomials of degree 2 with real coefficients. Any polynomial of degree 2n - 1 > 1 and real coefficients can be factored into a polynomial of degree 1 and n polynomials of degree 2. Polynomials of degree 2 (quadratics) with real coefficients may or may not be factorable in turn into polynomials of degree 1 with real coefficients.
Third, finding factorizations, even approximately, for polynomials of degree > 4 will usually require advanced methods. Indeed, factoring even polynomials of degree 3 or 4 may be arduous. If the coefficients are rational, you can use the rational or integer root theorems to find a "nice" factorization if there is one. In other words, factoring ranges from the obvious to the virtually impossible without a computer program.
Fourth, your specific problem involves a quadratic. You can factor any quadratic by using the quadratic formula although in some cases the factoring is more quickly done by reversing the FOIL method. However, the quadratic formula works every time and does not take that long.
[MATH]ax^2 + bx + c = a(x^2 + px + q) = a * \left (x + \dfrac{p + \sqrt{p^2 - 4q}}{2} \right ) \left (x + \dfrac{p - \sqrt{p^2 - 4q}}{2} \right ).[/MATH]
If [MATH]p^2 < 4q[/MATH], there is no factoring with real coefficients.
Finally, if you are trying to determine whether a rational function can be simplified and do not see a factorization, try dividing the polynomial of higher degree by the one of lower degree.
First, what you have provided is a polynomial expression rather than a rational expression.
Second, any polynomial of degree n > 0 can be factored into n polynomials of degree 1, perhaps by using complex coefficients. Any polynomial of degree 2n > 0 and real coefficients can be factored into n polynomials of degree 2 with real coefficients. Any polynomial of degree 2n - 1 > 1 and real coefficients can be factored into a polynomial of degree 1 and n polynomials of degree 2. Polynomials of degree 2 (quadratics) with real coefficients may or may not be factorable in turn into polynomials of degree 1 with real coefficients.
Third, finding factorizations, even approximately, for polynomials of degree > 4 will usually require advanced methods. Indeed, factoring even polynomials of degree 3 or 4 may be arduous. If the coefficients are rational, you can use the rational or integer root theorems to find a "nice" factorization if there is one. In other words, factoring ranges from the obvious to the virtually impossible without a computer program.
Fourth, your specific problem involves a quadratic. You can factor any quadratic by using the quadratic formula although in some cases the factoring is more quickly done by reversing the FOIL method. However, the quadratic formula works every time and does not take that long.
[MATH]ax^2 + bx + c = a(x^2 + px + q) = a * \left (x + \dfrac{p + \sqrt{p^2 - 4q}}{2} \right ) \left (x + \dfrac{p - \sqrt{p^2 - 4q}}{2} \right ).[/MATH]
If [MATH]p^2 < 4q[/MATH], there is no factoring with real coefficients.
Finally, if you are trying to determine whether a rational function can be simplified and do not see a factorization, try dividing the polynomial of higher degree by the one of lower degree.
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