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

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### 3 Answers

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`

Group:

`(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

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.

p+1=0

p=-1.

And:

p-6=0

p=6.

The two solutions to equation are -1 and 6.

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.