1, exercise p2 _ 2p _ 15 = 0 2, exercise (x _1) (x+2)=18 3, exercise (k _6) (k _ 1) = _ k _ 2 1, show and explain each step the process of factoring your quadratic. 2, show the use of the zero product properly in solving the quadratic . 3, be sure to check your solutions. 4 state your solution as a solution set

These are 3 questions and 3 answers:

Question 1) p^2 - 2p - 15 = 0

Solution:

1) factoring:

a) write two parenthesis with p as the first term (common term) of each:

(p    ) (p     )

b) in the first parenthesis copy the sign of the second term (i.e. - ), and in the second parenthesis put the product of the signs of the second and third terms (i.e. - * - = +)

=> (p -   )(p +   )

c) search two numbers whose sum is - 2 and its product is - 15. Those numbers are -5 and +3, look:

- 5 + 3 = - 2

( - 5)( + 3) = - 15

=> (p - 5)(p + 3)

d) you can prove that (p - 5) (p + 3) = p^2 - 20 - 15

2) equal the expression to zero and solve:

(p - 5)(p + 3) = 0 => (p - 5) = 0 or (p + 3) = 0

p - 5 = 0 => p = 5
p + 3 = 0 => p = - 3

=> that means that both p = 5 and p = - 3 are solutions.

3) check the solutions:

p = 5 => (5)^2 - 2(5) - 15 = 25 - 10 - 15 = 0 => check

p = - 3 => (-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0 => check

4) As a solution set: {5, - 3}

Question 2: (x - 1)(x + 2) = 18

Solution

0) expand, combine like terms and transpose 18:

x^2 + x - 2 - 18 = 0

x^2 + x - 20 = 0

1) factoring:

a) write two parenthesis with x as the first term (common term) of each:

(x    ) (x     )

b) in the first parenthesis copy the sign of the second term (i.e. + ), and in the second parenthesis put the product of the signs of the second and third terms (i.e. + * - = -)

=> (x +    )(x -   )

c) search two numbers whose sum is + 1 and its product is - 20. Those numbers are + 5 and - 4, look:

5 - 4 = 1

(5)( - 4) = - 20

=> (x + 5)(x - 4)

d) you can prove that (x + 5) (x - 4)) = x^2 + x - 20

2) equal the expression to zero and solve:

(x + 5)(x - 4) = 0 => (x + 5) = 0 or (x - 4) = 0

x + 5 = 0 => x = - 5
x - 4 = 0 => x = 4

=> that means that both x = - 5 and x =  4 are the solutions.

3) check the solutions:

x = - 5 => (- 5 - 1)(- 5 + 2) = 18

=> (-6)(-3) = 18

=> 18 = 18 => check

x = 4 => (4 - 1)(4 + 2) = (3)(6) = 18 => check

4) As a solution set: {- 5, 4}

Question 3: (k - 6) (k - 1) = - k - 2

Solution:

0) (k - 6)(k - 1) = - k - 2 =>

k^2 -7k + 6 = - k - 2 =>

k^2 -7x + k + 6 + 2 = 0 =>

k^2 - 6k + 8 = 0

1) factor

(k - ) (k - ) = 0

(k - 4) (k - 2) = 0 < - 4 - 2 = - 6 and (-4)(-2) = +8

2) Zero product property:

k - 4 = 0 , k - 2 = 0

=> k = 4, k = 2

3) Check:

a) (4 - 6) (4 - 1) = - 4 - 2

(-2)(3) = - 6

- 6 = - 6 => check

b) (2 - 6)(2 - 1) = - 2 - 2

(-4)(1) = -4

-4 = - 4 => check

4) Solution set: {2, 4}

The answer is a hoped that