`p^2 - 5p - 6` Factor the trinomial. If the trinomial cannot be factored, say so.

Textbook Question

Chapter 5, 5.2 - Problem 32 - McDougal Littell Algebra 2 (1st Edition, Ron Larson).
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atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

Try to see this problem as:

ax^2 + bx + c

` `

The first step is to find factors of c (-6) that add up to -5

2 x 3 = 6

and 1 x 6 = 6

we know that 6 -1 = 5 so that works.

These numbers would be 6 and -1, so we plug them in as b

`p^2 - p + 6p - 6`


`(p^2 - p) + (6p - 6)`

Factor out common factors:

p (p - 1 ) + 6 (p - 1)

Now set these parentheses equal to 0 and solve:

(p+6) (p-1)

p+6 = 0

p = -6

p-1 = 0

p = 1

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aishukul's profile pic

aishukul | Student, Grade 10 | (Level 3) Honors

Posted on

 We can use intercept form to factor easily. 

(p+1)(p-6)=0  We got this because we choose specific numbers that satisfy the conditions that the two numbers have to add up to -5 and also multiply to -6. 1+-6 equals -5 and 1*-6=-6 so the numbers that I factored are correct.

Now that we have this, we can find the solutions to the equation by moving the numbers to the right of equal sign.






The two solutions to equation are -1 and 6. 

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udonbutterfly's profile pic

udonbutterfly | Student, College Freshman | (Level 1) Valedictorian

Posted on

Usually for equations like this you do not need to go through the while entire steps since the leading coefficient has a 1. all you need to is the multiplies of the last in this case 6 that add up to the middle term, -5 and fill in the formula (p+/- no.)(p+/- no.) I will demonstrate.

So 6x1=6 1-6=-5

and 2x3=6 -2-3=-5

Now Here is the thing both of the multiples of 6, when added, equal -5 but there is only one right one and it's 1-6. This is the right answer because when multiplied 1x-6 will give you -6 whereas -2 x-3 will you a positive 6.

Not take 1-6 and fill in the blank

(p+1)(p-6) You can check your answer by foiling and you'll et the same answer.

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