factoring-polynomials

# You can check this by multiplying out the right side

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You can check this by multiplying out the right side. Example. Factor 12 x 2 + 11 x 5. I need numbers a and b whose product is 12, and such that 12 x 2 + 11 x 5 = ( ax + 5)( bx 1) or 12 x 2 + 11 x 5 = ( ax 5)( bx + 1) . If you try various combinations, you’ll find that 12 x 2 + 11 x 5 = (4 x + 5)(3 x 1) . Examples. Factoring quadratics is something you should be able to do fluently. Here are some more for practice. To factor x 2 + 9 x + 14, I need two numbers which add to 9 and multiply to 14. The pairs of numbers multiplying to 14 are (1 , 14) and (2 , 7). 2 + 7 = 9, so I use 2 and 7: x 2 + 9 x + 14 = ( x + 2)( x + 7) . 3

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To factor x 2 9 x + 18, I need two numbers which add to 9 and multiply to 18. The pairs of numbers which multiply to 18 are (1 , 18), ( 1 , 18), (2 , 9), ( 2 , 9), (3 , 6), and ( 3 , 6), ( 3) + ( 6) = 9, so I use 3 and 6: x 2 9 x + 18 = ( x 3)( x 6) . To factor x 2 14 x + 49, you can look for two numbers which add to 14 and multiply to 49. 7 and 7 work. I would do it differently: The perfect square 49 makes me think that maybe this is a standard form. 49 = 7 2 , and 2 · 7 = 14, which is the middle coefficient — and that led me to ( x 7) 2 . I used 7 because the middle coefficient was 14. So either way, x 2 14 x + 49 = ( x 7) 2 . To factor x 2 2 x 24, I need two numbers which add to 2 and multiply to 24. The pairs of numbers which multiply to 24 are (1 , 24), ( 1 , 24), (2 , 12), ( 2 , 12), (3 , 8), ( 3 , 8), (4 , 6), and ( 4 , 6). 4 + ( 6) = 2, so x 2 2 x 24 = ( x 6)( x + 4) . Don’t forget that you can always check your factoring by multiplication! To factor 4 x 2 + 4 x + 1, I notice that both the 4 in 4 x 2 and the 1 are perfect squares. 4 x 2 = (2 x ) 2 and 1 = 1 2 ; is this a standard form? Well, 2 · 2 x = 4 x , so 4 x 2 + 4 x + 1 = (2 x + 1) 2 . To factor 2 x 2 5 x + 3, I notice that 2 x 2 breaks down as 2 x · x and 3 could be either 1 · 3 or ( 1) · ( 3).
• Spring '14
• AaronW.Phillips

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