This process is usually used when the leading coefficient (the a term) is a number other than “1,” but it can also be used for quadratic equations in which a = 1. Example: 2x2 + 9x + 10
Example: 2x2 + 9x + 10 a = 2; c = 10 a * c = 2 * 10 = 20
Example: The factors of 20 are: 1, 2, 4, 5, 10, 20 Written in factor pairs: (1, 20), (2, 10), (4, 5)
If your master product was negative, you will need to find a pair of factors that equal the b term when subtracted from one another. Example: 2x2 + 9x + 10 b = 9 1 + 20 = 21; this is not the correct pair 2 + 10 = 12; this is not the correct pair 4 + 5 = 9; this is the correct pair
Note that the order of the center terms should not matter for this problem. No matter which order you write the terms in, the end result should be the same. Example: 2x2 + 9x + 10 = 2x2 + 5x + 4x + 10
Example: 2x2 + 5x + 4x + 10 = (2x2 + 5x) + (4x + 10)
Example: x(2x + 5) + 2(2x + 5)
Example: (2x + 5)(x + 2)
Example: 2x2 + 9x + 10 = (2x + 5)(x + 2) The final answer is: (2x + 5)(x + 2)
a * c = 4 * -10 = -40 Factors of 40: (1, 40), (2, 20), (4, 10), (5, 8) Correct factor pair: (5, 8); 5 - 8 = -3 4x2 - 8x + 5x - 10 (4x2 - 8x) + (5x - 10) 4x(x - 2) + 5(x - 2) (x - 2)(4x + 5)
a * c = 8 * -3 = -24 Factors of 24: (1, 24), (2, 12), (4, 6) Correct factor pair: (4, 6); 6 - 4 = 2 8x2 + 6x - 4x - 3 (8x2 + 6x) - (4x + 3) 2x(4x + 3) - 1(4x + 3) (4x + 3)(2x - 1)
Usually, you will use this method when you see a polynomial equation that looks like: ax3 + bx2 + cx + d The equation may also look like: axy + by + cx + d ax2 + bx + cxy + dy ax4 + bx3 + cx2 + dx Or similar variations. Example: 4x4 + 12x3 + 6x2 + 18x
If the only thing all four terms has in common is the number “1,” there is no GCF and nothing can be factored out at this point. When you do factor out a GCF, make sure that you continue to keep it at the front of your equation as you work. This factored out GCF must be included as part of your final answer for that answer to be accurate. Example: 4x4 + 12x3 + 6x2 + 18x Each term has 2x in common, so the problem can be rewritten as: 2x(2x3 + 6x2 + 3x + 9)
If the first term of the second group has a minus sign in front of it, you will need to put a minus sign in front of the second parentheses. You will need to change the sign of the second term in that grouping to reflect that choice. Example: 2x(2x3 + 6x2 + 3x + 9) = 2x[(2x3 + 6x2) + (3x + 9)]
At this point, you might be faced with a choice between factoring out a positive number or a negative number for the second group. Look at the signs before the second and fourth terms. When the two signs are the same (both positive or both negative), factor out a positive number. When the two signs are different (one negative and one positive), factor out a negative number. Example: 2x[(2x3 + 6x2) + (3x + 9)] = 2x2[2x2(x + 3) + 3(x + 3)]
If the binomials inside the current sets of parentheses do not match, double-check your work or try rearranging your terms and grouping the equation again. The parentheses must match. If they do not match no matter what you try, the problem cannot be factored by grouping or by any other method. Example: 2x2[2x2(x + 3) + 3(x + 3)] = 2x2[(x + 3)(2x2 + 3)]
Example: 4x4 + 12x3 + 6x2 + 18x = 2x2(x + 3)(2x2 + 3) The final answer is: 2x2(x + 3)(2x2 + 3)
2[3x2 + xy - 12x - 4y] 2[(3x2 + xy) - (12x + 4y)] 2[x(3x + y) - 4(3x + y)] 2[(3x + y)(x - 4)] 2(3x + y)(x – 4)
(x3 - 2x2) + (5x - 10) x2(x - 2) + 5(x - 2) (x - 2)(x2 + 5)
