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5.7-5.9 More about Factoring Trinomials. To factor ax 2 +bx+c when a≠1 find the integers k,l,m,n such that ax 2 +bx+c=( k x+ m )( l x+ n ) = kl x 2 +( kn+lm ) x+ mn Therefore k and l must be factors of a m and n must be factors of c
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To factor ax2+bx+c when a≠1 find the integers k,l,m,n such that ax2+bx+c=(kx+m)(lx+n) = klx2+(kn+lm)x+mn Therefore k and l must be factors of a m and n must be factors of c The goal is to find a combination of factors of a and c such that the outer and inner products add up to the middle term bx. Factoring a trinomial of the form ax2+bx+c
3x2 + x -4 • The possible factors for the first term are 3,1 or 1,3. The possible factors of the last term are 2, -2; -2, 2; 4, -1; and -4, 1. Here are all the possible factors and what the middle term equals • (3x -2) (x + 2) = 4x for the middle term • (3x + 2) (x – 2) = -4x for the middle term • (3x -1) (x + 4) = 11x for the middle term • (3x-4) (x + 1) = -x for the middle term • (3x + 1) (x-4) = -11 for the middle term • (3x +4) (x – 1) = x for the middle term ANSWER Example
1) • 2) • 3) Examples:
Factoring Completely • 4) • 5) Examples:
Factor: 2x2+ x -15 • In order to ‘group’ this polynomial, we actually have to ‘break out’ the middle term. • In order to figure out what coefficients to use when we break out the middle term, we need to choose factors of the product ac that add up to b. • In the example above the product of ac is -30 (-15*2). The factors of -30 are +/- 1, 30, 5, 6, 3 10, 2 and 15. Looking at this factors, we see that -5 + 6 will equal one. • So now we can write out this polynomial like this: • 2x2 + 6x-5x -15. • 2x(x + 3) -5(x+3) • (2x-5)(x+3) Factoring by grouping
If necessary, write the trinomial in standard form • Choose factors of the product ac that add up to b • Use these factors to rewrite the middle term as a sum. • Group factor and solve. Factoring by grouping
1) • 2) Examples by grouping: