Respuesta :
(p - 5)(2p - 5), in the first option
Further explanation
The Problem:
Which are the factors of 2p² - 15p + 25 = ?
- (p - 5)(2p - 5)
- (p - 5)(2p + 5)
- (p + 5)(2p - 5)
- (2p - 15)(p + 5)
The Process:
The problem we have this time is how to solve factoring a quadratic equation.
First Way
2p² - 15p + 25 = 2p² - 10p - 5p + 25
= 2p(p - 5) - 5(p - 5)
= (2p - 5)(p - 5) or (p - 5)(2p - 5)
Recall this: [tex]\boxed{ \ a(p + q) + b(p + q) \rightleftharpoons (a + b)(p + q) \ }[/tex]
Second Way
Allow me to introduce the "temporary quadratic" method. This is a unique way! Look carefully below.
2p² - 15p + 25 ⇒ multiply 2 as the coefficient of the p² term with 25 as a constant.
2 x 25 = 50.
Then, find two factors of 50 which if added are -15 as the coefficients of the term p.
We get the following: - 5 x - 10 = 50, i.e., -5 and -10.
Both are positioned in factor brackets. Simultaneously we place the term of 2p in both brackets as a sum operation with both factors. This is properly the uniqueness of this method!
2p² - 15p + 25 = "(2p - 10)(2p - 5)"
Because we call it the "temporary quadratic" method, simplify the factors in parentheses if it can still be done.
Thus, we get 2p² - 15p + 25 = (p - 5)(2p - 5), i.e. the two terms of (2p - 10) divided by 2.
Third Way
Let us check one by one of the answer options available.
(p - 5)(2p - 5) = p(2p - 5) - 5(2p - 5)
= 2p² - 5p - 10p + 25
= 2p² - 15p + 25 (yes, this is correct)
(p - 5)(2p + 5) = p(2p - 5) - 5(2p + 5)
= 2p² - 5p - 10p - 25
= 2p² - 15p - 25
(p + 5)(2p - 5) = p(2p - 5) + 5(2p - 5)
= 2p² - 5p + 10p + 25
= 2p² + 5p + 25
(2p - 15)(p + 5) = 2p(p + 5) - 15(p + 5)
= 2p² + 10p - 15p - 75
= 2p² - 5p - 75
_ _ _ _ _ _ _ _ _ _
Take a look at this example:
3m² - 10m - 8 = ?
- = 3m² - 12m + 2m - 8
- = 3m(m - 4) + 2(m - 4)
- This becomes (3m + 2)(m - 4)
3m² - 10m - 8 = ?
- 3 x (-8) = -24
- -24 = 2 x (-12), because 2 + (-12) = -10
- 3m² - 10m - 8 = "(3m + 2)(3m - 12)"
- This becomes (3m + 2)(m - 4)
Notes:
[tex]\boxed{ \ ax^2 + bx + c = 0, \ \ \ \ \ a \neq 0 \ }[/tex] is a quadratic equation in standard form.
- x is a variable
- a, b, and c are real number coefficients
- the only c is called a constant (no variables)
Methods of solving quadratic equations include:
- Factoring and using the zero product property:
- Using the square root property: [tex]\boxed{ \ x^2 = a, \ then \ x = \pm \sqrt{a} \ }[/tex]
- Completing the square: [tex]\boxed{ \ x^2 + bx + \bigg({\frac{b}{2}} \bigg)^2 = \bigg( x + \frac{b}{2} \bigg)^2 \ }[/tex]
- Using the quadratic formula: [tex]\boxed{ \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ }[/tex]
Learn more
- Factoring methods https://brainly.com/question/2568692
- A word problem of a quadratic equation https://brainly.com/question/11805547
- What are the coordinates of the vertex of the graph? https://brainly.com/question/1286775
