Question

Can someone please explain to me how to factor this s^2+13s+40?

Answer 1

ok

Answer 2

will u give fan and medal

Answer 3

yes if you help me

Answer 4

You have a trinomial of the form:

\(x^2 + ax + b\)

The variable in your case is s instead of x, but the form is the same.
The important thing to see is that the square variable is only multiplied by 1.

Answer 5

8+5=13
8*5=40

Answer 6

All you need to do is set up two sets of parentheses.

\((s~~~~~)(s~~~~~~~)\)

Then you need to find two numbers whose product is b and whose sum is a.

Answer 7

If you compare your polynomial with
\(x^2 + ax + b\)
\(s^2 + 13s + 40\)

you have a = 13 and b = 40

What are two numbers that multiply to 40 and add to 13?

Answer 8

40 =
40 * 1 = 40; 40 + 1 = 41 (not 13)
20 * 2 = 40; 20 + 2 = 22 (not 13)
10 * 4 = 40; 10 + 4 = 14 (not 13)
8 * 5 = 40; 8 + 5 = 13 (just what we need)

That means the two numbers you need that multiply to 40 and add to 13 are 8 and 5.

You already had the two sets of parentheses with s in them, now just write each of the numbers in a parentheses:

\(s^2 + 13s + 40 = \)

\(=(s~~~~~)(s~~~~~~) \)

\(=(s + 8)(s + 5) \)

Answer 9

Keep in mind that this method only works for a square trinomial of the form
\(x^2 + ax + b\)
where the x^2 term has no coefficient (other than 1).

Answer 10

@riley101 Do you understand?

Answer 11

Would you like to see another example?

Answer 12

I think i understand but would it be different if it was subtraction instead of addition?

Answer 13

so like 2s^2-14s-156?

Answer 14

The method is still the same.
Let me do this simpler example.
Then I'll show you how to do yours.

Here is an example where there is a subtraction.

Factor:
\(x^2 -2x - 15\)

Answer 15

Here you have again a trinomial of second degree.
You see an x^2 term with no number multiplying it.
It is followed by an x-term and by a number.
It follows the form
\(x^2 + ax + b\),
so we can use our method.

1. Setup two sets of parentheses with just x on the left side

\(x^2 - 2x - 15\)

\(= (x~~~~~~~)(x~~~~~~~)\)

Answer 16

Now you need two numbers that multiply to -15 and add to -2.

Answer 17

Since -15 is a negative number, the only two numbers that multiply to a negative number are a positive number and a negative number.

What two numbers multiply to -15 and add to -2?

Answer 18

whats the easiest way to find those numbers without having to divide starting from one going up?

Answer 19

5-3?

Answer 20

or -3-5?

Answer 21

5 and -3 do multiply to -15.
Now add them. What do you get?

Answer 22

2?

Answer 23

but its not negative

Answer 24

Yes.
We need the sum to be -2, not positive 2.
All you need is a small adjustment.
Instead of 5 and -3, you need -5 and 3.

Answer 25

okay so (x+-5)(x+3)

Answer 26

Now look:
-5 * 3 = -15; -5 + 3 = -2
Now we get the product of -15 and the sum of -2.
That confirms that the numbers we need are -5 and 3.

Now you just place each number inside a set of parentheses following the x.

\((x - 5)(x + 3)\)

Answer 27

You are correct.
It's just that x + (-5) is simply x - 5.

Answer 28

Now we can try your problem.
Are you ready?

Answer 29

Okay thank you. I'm ready

Answer 30

Factor

\(2s^2-14s-156\)

Answer 31

The first rule of factoring:
Try to factor a common factor from all terms if possible.

Answer 32

Do the numbers 2, -14, and 156 have a common factor?

Answer 33

the common factor is 2

Answer 34

Correct.
We first factor a 2 from all terms.
Can you do that?

Answer 35

-7, 78

Answer 36

is it -156, actually?

Answer 37

When we factor out a common factor, we are using the distributive property in reverse.
The result of factoring out a common term must be a factored expression that when you use the distributive property on it, you get back the original expression.

\(2s^2-14s-156\)

\(2(s^2 - 7s - 78)\)

Answer 38

Notice that if you were to start with the second line above and distribute the 2, you'd end up with your original problem.

Answer 39

Okay thats what I had initially and i thought that was the final answer lol

Answer 40

Now that we have factored out the 2, we end up with a second degree trinomial that starts with simply \(s^2\). That is the kind of trinomial we have been factoring in this post, so we can do it.

Answer 41

Now we ask the question:
What two numbers multiply to -78 and add to -7?

Answer 42

The thing i am having the hardest time with is finding numbers that are the sum and product of the negative numbers

Answer 43

Start by finding any numbers that multiply to 78.
Just think of any two positive numbers that multiply to 78.
One pair is always easy: 78 and 1.
Can you think of other pairs of numbers whose product is 78?

Answer 44

What is 78 divisible by?

Answer 45

6 x13

Answer 46

-13 +6

Answer 47

Excellent.
6 * 13 = 78.
Since we really need the product to be -78, and not 78, we need to have a positive number and a negative number.

You can have
-6 * 13
or 6 * (-13)
Both have a product of -78 that we need.
Does either one of those two pairs add to -7?

Answer 48

Oh, you got it already.
You're too fast for me.

Answer 49

Great.
The numbers are -13 and 6 since
-13 * 6 = -78
-13 + 6 = -7

Answer 50

so its 2(s-13)(s+6)?

Answer 51

Now all you need to do is place the two numbers you found int he parentheses like we did above.

\(2s^2-14s-156\)

\(2(s^2 - 7s - 78)\)

\(2(s - 13)(s + 6) \)

Answer 52

Correct.
Once again, you beat me to it.
Next time, I'll ask you for help!

Answer 53

THANK YOU SO MUCH!!!

Answer 54

Ok, sorry, but gtg.

Answer 55

BTW, this method works for second degree trinomials in which the coefficient of the square term is 1. If the coefficient of the square term is not 1, there is a different method that is a little more complicated, but it's still pretty simple to do.

For example: \(6x^2 - 11x - 35\) cannot be factored using this method.
6, -11, and -35 have no common factors, so you can't factor out a common factor to make this trinomial start with a coefficient of 1.