Factor Diamond Practice Problems


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1 Factor Diamond Practice Problems 1. x 2 + 5x x 2 +7x x 2 + 9x x 2 + 9x x 2 7x x 2 x x 2 + x y 2 x x + 6x x 2x 2
2 Factor Diamond Practice Problems Walkthrough 1. x 2 + 5x Put the 6 on the top and the 5 on the bottom (x + 3)(x + 2) Now that you have the equation find the number which add or subtracts to make a 0 Lets start with the first one, (x + 3) a 3 would make it 0, (3 + 3) = 0 For (x + 2) it would be a 2, (2 + 2) = 0 So your two answers are 2. x 2 +7x Put the 12 on top and the 7 on the bottom Now its (x + 3)(x + 4), make them 0 For (x + 3) it would be 3, (3 + 3) = 0 For (x + 4) it would be 4 (4 + 4) = 0 So your two answers would be x 2 + 9x + 8 Put the 8 on top and the 9 on bottom Now its (x + 1)(x + 8) For (x + 1) its 1, (1 + 1) = 0 For (x + 8) its 8, (8 + 8) = 0 So your two answers would be
3 4. x 2 + 9x +14 Put the 14 on top and the 9 on bottom Now its (x + 2)(x + 7) For (x + 2) its 2, (2 + 2) = 0 For (x + 7) its 7, (7 + 7) = 0 So your two answers are 5. 2x 2 7x 4 Here you will be using the borrowing method, so multiply the first and last numbers, so the equation turns into x 2 7x 8 Now plug it into the diamond, the 8 on top and the 7 on bottom  Now its (x  8)(x + 1) But since you borrowed a 2 you have to give it back, so divide both numbers by 2  For (x  8) it would turn into (x  8/2)  For (x + 1) it would turn into (x + ½) And reduce them both  8/2 reduces to 4, and ½ can t reduce any further so the equations are (x 4)(x + ½)  To get rid of the denominator multiply by it, in this problem it is (x + ½) so when multiplied by 2, it turns into (2x + 1)  Our final two equations are (x 4)(2x + 1) and make them both 0  For (x 4) it would be 4, (4 4) = 0  For (2x + 1) it would be ½, because 2 times ½ = 1, and (1 + 1) = 0  So your two final answers are So your two answers are
4 6. 3x 2 x 4 Multiply the first and last numbers, so the equation turns into x 2 x 12 Now put it into the diamond, 12 on top and the 1 on bottom  Now its (x 4)(x + 3) But you have to give back the 3 so (x 4/3)(x + 3/3)  Simplify them and if they cant be simplified anymore multiply by the denominator So (x 4/3) cant be simplified so you multiply by 3 which turns the problem into (3x 4) For (x + 3/3) simplifies to (x + 1)  For (3x 4) it would be 4/3, (44) = 0  For (x + 1) it would be 1, (1 + 1) = 0  So your two answers are 7. 5x 2 + x 18 Multiply the first and last numbers, so the equation turns into x 2 + x  90 Now put it into the diamond, 90 on top and 1 on bottom  Now its (x 9)(x + 10) But you have to give back the 5 so (x 9/5)(x + 10/5) So (x 9/5) cant be simplified so you multiply by 5 which turns the problem into (5x 9) For (x + 10/5) simplifies to (x + 2)  For (5x 9) it would be 9/5, (99) = 0  For (x + 2) it would be 2, (2 + 2) = 0 So your final two answers are
5 8. 2y 2 y 1 Multiply the first and last numbers, so the equation turns into y 2  y  2 Now put it into the diamond, 2 on top and 1 on bottom  Now its (x 2)(x + 1) But you have to give back the 2 so (x 2/2)(x + 1/2) So (x 1/2) cant be simplified so you multiply by 2 which turns the problem into (2x 1) For (x + 2/2) simplifies to (x + 1)  For (2x + 1) it would be 1/2, (1 + 1) = 0  For (x  1) it would be 1, (1 1) = 0 So your two solutions are x + 6x 2 First you have to put it in descending order which changed the equation to 6x 213x + 6 and use the borrowing method  This changed the problem to x 213x + 36 Now put it into the diamond, 36 on top and 13 on bottom  Now its (x 9)(x  4) But you have to give back the 6 so (x 9/6)(x  4/6) So (x 9/6) simplifies to 3/2 and it cant be simplified anymore so you multiply by 2 which turns the problem into (2x 3) When simplifying (x  4/6) turns to (x  2/3) and now you multiply by 3. This turns the problem into (3x  2) For (2x  3) it would be 3/2, (3 3) = 0  For (3x  2) it would be 2/3, (2 + 2) = 0 The two final answers are
6 x 2x 2 First you have to put it in descending order which changed the equation to 2x 2 + x + 15 and use the borrowing method  This changed the problem to x 2 + x  30 Now put it into the diamond, 30 on top and 1 on bottom  Now its (x 5)(x + 6) But you have to give back the 2 so (x 5/2)(x + 6/2) So (x 6/2) simplifies to 3 which turns the problem into (x 3) (x 5/2) cannot be simplified anymore so you multiply it by 2 which turns it into (2x  5). For (x  3) it would be 3, (3 3) = 0  For (2x  5) it would be 5/2, (55) = 0 The two final answers are
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This booklet outlines the methods we teach pupils for place value, times tables, addition, subtraction, multiplication, division, fractions, decimals, percentages, negative numbers and basic algebra Any
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