ax 2 + bx + c 3x 2 + 7x – 6 ac = 3 × – 6 = – 18 b = 7 the two numbers are 9 and – 2 Split the middle term bx into two like terms 3x 2 + 9x – 2x – 6 Factorise by grouping the four terms into pairs 3x (x + 3) – 2(x + 3) (x + 3) (3x – 2) Therefore 3x 2 + 7x – 6 = (3x – 2) (x + 3) Example 2 Factorise : – 11x + 7x 2 – 6 Solution 1) Write the expression in the form ax 2 + bx + c 7x 2 – 11x – 6 a × c = 7 × – 6 = -42 2) Find two numbers that both multiply to -42 and add to – 11. Solution We require two numbers that multiply to – 18 and add to 7. Factorise by grouping the four terms into pairs.split the middle term bx into two like terms using those two numbers as coefficients.
find two numbers that both multiply to ac and add to b.write the expression in the form ax 2 + bx + c.Split the 11x into 8x + 3x 2x 2 + 8x + 3x + 12 2x (x + 4) + 3 (x + 4) (x + 4) (2x + 3) To factorise a quadratic trinomial We need two numbers with a product of 24 and a sum of 11. So to factorise 2 × 12 = 24 2x 2 + 11x + 12 Numbers Looking at line (3) and (4): a × c = 2 × 12 = 24 = (which is 8×3) b = 11 = 8 × 3 So the key is that the split – up of bx to obtain the like terms 8x + 3x in (2) comes from the pattern: ac (coefficient of x 2 times constant term) gives the product of the required two numbers: b (coefficient of x) gives their sum. (2) = 2x 2 + (8 + 3)x + 12……………………… (3) The factorisation pattern is no longer so obvious. Use Eisenstein’s Criterion, which is described in the Tips section, for non-quadratic trinomials. When all the answers are the square root of a negative number, there are no real solutions, so there are no factors. Now let us adapt our basic procedures to quadratics with coefficients of x 2 other than 1. When you’re stuck on quadratic trinomials ((ax2+bx+c)), use the quadratic formula. In all previous cases, the coefficient of x 2 has been 1, producing simple quadratics. Non – monic quadratic have coefficients of x 2 greater than 1.
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