Special pattern: difference of two perfect squares

Special pattern: perfect square trinomials

Factoring [math]x^2+bx+c[/math] - when the leading coefficient is 1.

When the leading coefficient is not 1: the most general case, [math]ax^2+bx+c[/math].[br][br]When I try to factor a trinomial like [math]x^2+3x+5[/math], I ask myself, "What is a pair of numbers that multiply to give me 5, and add to give me 3?" There is no such pair of real numbers, so I know that this trinomial can't be factored. We'll learn to use the quadratic formula to find its roots. In fact, we'll learn that they are a pair of complex conjugates - more on that later.[br][br]But now it's time to look at more general quadratics, what do we do when the leading coefficient isn't 1?[br][br][br]First check for common factors. [math]3x^2+21x+30[/math] looks hard, but if we notice that 3 is a factor of every term, we can turn this into a simpler problem by factoring it out: [math]3x^2+21x+30=3\left(x^2+7x+10\right)[/math]. Since 5•2 =10 and 5+2=7, I have [math]3\left(x+2\right)\left(x+5\right)[/math] as my factored form.[br][br]If the leading coefficient is negative, I always factor out a -1 first.[br][br]Let's look at and example that doesn't have a common factor: [math]6x^2+19x+10[/math]. If this can be factored into two binomials, we know that the first term of the trinomial comes from multiplying the first terms of the binomials. The factors could be [math]\left(6x+?\right)\left(x+?\right)[/math] or [math]\left(3x+?\right)\left(2x+?\right)[/math] or [math]\left(2x+?\right)\left(3x+?\right)[/math]. The 10 is going to come from multiplying the second terms of the two binomials, so the ?'s could be 10 and 1, 1 and10, 2 and 5, or 5 and 2. That's a lot of possibilities we need to try to find the one combination that gives us 19[i]x[/i] for the middle term.[br][br]Fortunately, there is a simple trick we can use so that we don't have to guess and check. We start by multiplying the first and last coefficients. In our case, 6 • 10 = 60. Now we look for a pair of numbers that multiply to give us this number and add to give us 19.[br][br]60 | [br] 60, 1 61[br] 30,2 32[br] 20,3 23[br] 15,4 19[br] 12,5 17[br] 10,6 16[br][br]15 and 4 are the factors we need.[br][br]Now we write our trinomial as a quadranomial by splitting the middle term into these two numbers:[br][br] [math]6x^2+19x+10=[/math] [math]6x^2+15x+4x+10[/math][br][br]To finish, we are going to "factor by grouping." We'll look at the first two terms and the last two terms and factor out any common factors from both:[br][br] [math]6x^2+15x+4x+10=[/math] [math]\left(6x^2+15x\right)+\left(4x+10\right)[/math][br]       [math]3x\left(x+5\right)+2\left(x+5\right)[/math][br][br]In this method, we will always have the same binomial. We now factor out that common binomial, and we have factored our trinomial completely:[br][br][math]3x\left(x+5\right)+2\left(x+5\right)=\left(x+5\right)\left(3x+2\right)[/math][br][br]We can quickly check our work by multiplying the two binomials to see that we get our original trinomial.

Here are some examples for you to try:[br][math]3x^2+10x-8[/math][br][math]4x^2+19x+12[/math][br][math]10x^2-39x+14[/math][br][math]3x^2+20x+12[/math][br][math]18x^2+15x-25[/math][br][math]24x^2+26x-15[/math]