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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Factoring Trinomials

## Factoring Quadratic Trinomials

 WHAT TO DO: HOW TO DO IT: Given general trinomial of type that has no common factor. Read the "clues of the signs".[Read Â± as + or - ] The constant c is the grouping number GN = c Find all possible factors of GN = c whose sum or difference is b (depending on the sign before c .) + sum or - difference c = r Â· s , r > s (r + s) = b (r - s) = b 1. Consider the trinomial : x 2 - 5x + 6 The leading coefficient is 1 and the last sign is + The factors of 6 with a sum of 5 are 3 and 2 Since the last sign is + the "same sign" in both binomials is - (the sign of the middle term). Factor by grouping. Bring down the middle sign. Complete the factors. 1. x 2 - 5x + 6 r Â· s = 6 and r + s = 5 2. First, examine the trinomial to see if it has a common factor(s) in each term. Check the remaining trinomial to see if it will factor: 2. 3x 2 +15x + 123(x 2 + 5x + 4) 2a) Consider the trinomial : x 2 + 5x + 4 The leading coefficient is 1 and the last sign is + The factors of 4 with a sum of 5 are 4 and 1 Since the last sign is + the "same sign" in both binomials is + (the sign of the middle term). Factor by grouping. Bring down the middle sign. Complete the factors. 2a) x 2 + 5x + 4 r Â· s = 4 and r + s = 5 2b) Go back to the polynomial in the previous step: Replace the polynomial with the factors to find all of the factors: 3x 2 +15x + 12 2b) 3(x 2 + 5x + 4) = 3(x + 1)(x + 4) 3. Consider the trinomial with last sign + Two numbers with a product of 14 and sum of are 7 and 2 Rewriting the terms to factor by grouping: 3. x 2 + 9x + 14x 2 + 7x + 2x + 14 Then group the terms 2 Ã— 2 into two terms. Factor common factors. Bring down the middle sign. Find the common factor for each group. Factor out the common factor (x + 7)(x + 2) Answer Check by multiplying back out by F 0 I L FOIL method. First - Outer - Inner - Last Note O-I terms.

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