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

Question

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

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    2021-09-10T04:59:03+00:00

    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}

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