Factoring 8 - t^3Date: 01/07/98 at 19:53:11 From: Adam Jones Subject: Sum and difference of cubes My question is how to do this problem: Factor the given expression completely: 8 - t^3. Date: 01/12/98 at 11:51:56 From: Doctor Joe Subject: Re: Sum and difference of cubes Dear Adam, Let's start by looking at the following algebraic identity: x^3 - y^3 = (x - y)(x^2 + xy + y^2) To see that this is true, we start by expanding the expression on the right: (x - y)(x^2 + xy + y^2) = x(x^2 + xy + y^2) - y(x^2 + xy + y^2) = x^3 + (x^2)*y + xy^2 - (x^2)*y -xy^2 - y^3 = x^3 - y^3 Seeing that 8 = 2^3, we write x = 2 and y = t. Then, we have: 8 - t^3 = (2)^3 - (t)^3 = (2 - t)(2^2 + 2t + t^2) = (2 - t)(4 + 2t + t^2) But at this stage, you must be wondering: "How did you know that (x^3 - y^3) has a factor (x - y)?" Let's explore the real reason behind this clever guess. The guess lies in the following theorem (without it you would think I was very lucky). Theorem (Remainder/Factor Theorem) Let f(t) and d(t) be two polynomials such that the deg(f(t)) < deg(d(t)). Polynomials are things that look like this: f(t) = 2t^3 + 4t g(t) = -2t^7 + 4t^6 - 9t^5 + 40) and deg(polynomial) (read as the degree of polynomial) denotes the highest power of the indeterminate occurring in the polynomial, e.g.: deg (2t) = 1, deg (-4t^2 + 5t + 5) = 2) Then there exists a polynomial q(x) and r(x) such that deg(r(x))< deg(d(x)) or r(x) is identically equal to zero and f(t) = q(t)d(t) + r(t). *** End of Theorem *** Now, to apply this theorem (without proof) to this problem, we let f(t) = 8 - t^3. Observe that f(2) = 8 - 2^3 = 8 - 8 = 0. Let's divide f(t) by (t - 2). According to the Theorem above, there exist a q(t) and r(t) such that deg(r(t))< deg(t-2) = 1 (meaning that r(t) is actually a constant number, denoted by r) with f(t) = 8 - t^3 = (t- 2)q(t) + r The above statement is really an identity; i.e. the equality holds for all values of t. In this case we choose t = 2 - now you know why (t -2) is chosen - f(2) = 0 = (2 - 2)q(2) + r implying that r = f(2) = 0. So, (8 - t^3) leaves no remainder when it is divided by (t -2). Take note of the argument: there is no stage where long division is actually carried out to test whether (t -2) is a factor of f(t)! Now we know that 8 - t^3 = (t - 2)q(t) for some q(t). We apply the long division algorithm to find out the explicit form for q(t): -t^2 - 2*t - 4 ------------------------ t - 2 | -t^3 + 0*t^2 + 0*t + 8 -t^3 + 2*t^2 ------------ -2*t^2 + 0*t -2*t^2 + 4*t ------------ -4*t + 8 -4*t + 8 -------- 0 -------- Hence, 8 - t^3 = (t - 2)(-t^2 - 2t - 4) = (2 - t)(t^2 + 2t + 4) as desired. Now, test yourself by factorising completely 27 + t^3. All the best. -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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