Solutions to Problem Set 13

Math 101-06

10-6-2017

[Factoring]

1. Factor $24 x + 40$ .

$$24 x + 40 = 8 (3 x + 5).\quad\halmos$$


2. Factor $24 x^2 - 14 x$ .

$$24 x^2 - 14 x = 2 x (12 x - 7).\quad\halmos$$


3. Factor $2 x^3 y - 6 x y^2$ .

$$2 x^3 y - 6 x y^2 = 2 x y (x^2 - 3 y).\quad\halmos$$


4. Factor $x^2 + 6 x + 8$ .

I need two numbers whose product is 8 and whose sum is 6. The numbers are 2 and 4, so

$$x^2 + 6 x + 8 = (x + 2)(x + 4).\quad\halmos$$


5. Factor $x^2 + 12 x + 11$ .

I need two numbers whose product is 11 and whose sum is 12. The numbers are 1 and 11, so

$$x^2 + 12 x + 11 = (x + 1)(x + 11).\quad\halmos$$


6. Factor $x^2 - 5 x + 6$ .

I need two numbers whose product is 6 and whose sum is -5. The numbers are -2 and -3, so

$$x^2 - 5 x + 6 = (x - 2)(x - 3).\quad\halmos$$


7. Factor $x^2 - 10 x + 24$ .

I need two numbers whose product is 24 and whose sum is -10. The numbers are -4 and -6, so

$$x^2 - 10 x + 24 = (x - 4)(x - 6).\quad\halmos$$


8. Factor $x^2 - 6 x - 27$ .

I need two numbers whose product is -27 and whose sum is -6. The numbers are -9 and 3, so

$$x^2 - 6 x - 27 = (x - 9)(x + 3).\quad\halmos$$


9. Factor $x^2 - 12 x + 20$ .

I need two numbers whose product is 20 and whose sum is -12. The numbers are -2 and -10, so

$$x^2 - 12 x + 20 = (x - 2)(x - 10).\quad\halmos$$


10. Factor $x^2 + 7 x - 8$ .

$$x^2 + 7 x - 8 = (x + 8)(x - 1).\quad\halmos$$


11. Factor $2 x^3 + 6 x^2 + 4
   x$ .

Take out a common factor first, then factor the quadratic by trial:

$$2 x^3 + 6 x^2 + 4 x = 2 x(x^2 + 6 x + 2) = 2 x (x + 1)(x + 2).\quad\halmos$$


12. Factor $x^2 - 36$ .

$$x^2 - 36 = (x + 6)(x - 6).\quad\halmos$$


Let us train our minds to desire what the situation demands. - Seneca


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga