Question

I was trying to factor:

f(t) = -16t^2 - 32t + 128

I got (-16 - 16)(t - 1)(t + 8). What did I do wrong?

Answer 1

The first thing I did was split the middle term in half, so that I could factor by grouping.

Answer 2

@mathmale @mathstudent55 @boldjon @ganeshie8 @sleepyjess @triciaal

Answer 3

f(t) = -16t^2 - 32t + 128 = f(t) = -16t^2 - 16t - 16 + 128

Then i put them into groups

f(t) = (-16t^2 - 16t) - (16 + 128)

Answer 4

wait hold on for a minte

Answer 5

You could factor out the 16. That would simplify it enough.

Answer 6

the next step that I did was change the + sign to a minus

Answer 7

f(t) = (-16t^2 - 16t) + (-16 - 128)

Answer 8

The GCF in the first equation is -16t^2 and in the second binomial it is -16

Answer 9

So i divided the gcf out

Answer 10

f(t) = -16t(t - 1) + -16(t - 128)

Answer 11

Then I combined the terms

(-16t - 16)(t - 1)(t - 8)

Answer 12

First, you can't split the middle term. It only has one variable, which means it can't be split. To split, each would have to take a variable or have no variable.

Answer 13

oh

Answer 14

hold on, then should i split the second term?

Answer 15

Going back to what I said, just factor out a -16.

Answer 16

Don't split any terms. This other way is super simple. You are over complicating this.

Answer 17

alright

Answer 18

f(t) = -16t2 - 32t + 128

/ -16

f(t) = -16(t2 - 2t - 8)

Answer 19

that all?

Answer 20

It would be +2t

Answer 21

oh right

Answer 22

but then it's done?

Answer 23

Then just simplify what is in parenthesis.

Answer 24

ermm... how?

Answer 25

Do you know what FOIL is?

Answer 26

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Answer 27

yes, i do

Answer 28

@marihelenh Can we hurry please? I'm in a serious rush lol

Answer 29

Yeah, it is pretty much just the reverse. You just have to solve for the equation.

Answer 30

A rush?

Answer 31

how do I wait wut? i dont ge it

Answer 32

What are you in a rush for?

Answer 33

unspecified, please continue :)

Answer 34

Well, I actually have to go to. Do you want me to explain later or just the answer?

Answer 35

NO HELP HELP HELP HELP HELP PLEASE!

Answer 36

DONT GO, STAY FOR 5 MORE MINUTES PLEASE!

Answer 37

The first rule of factoring is to try to factor a common factor.

\(f(t) = -16t^2 - 32t + 128 \)

We see the coefficients are -16, 32, and 128.
They are all divisible by 16. Let's factor out -16 from all terms:

\(f(t) = -16(t^2 +2t -8) \)

Answer 38

He's got this! Sorry, but dinner's waiting!

Answer 39

Now we need to factor the trinomial \(t^2 + 2t - 8\)
It is a quadratic trinomial with a leading coefficient (the coefficient of the \(t^2\) term) of 1.

Answer 40

@marihelenh alright, that's fine

Answer 41

@mathstudent55 Yeah, I get it so far...

Answer 42

Notice the numbers in blue and red.

\(f(t) = -16(t^2 +\color{blue}{2}t \color{red}{-8})\)

We split the trinomial into two binomials:

\(f(t) = -16(t~~~~~~)(t~~~~~~)\)

Now to complete the binomials,
we need two numbers that multiply to \(\color{red}{-8}\) and add to \(\color{blue}{2}\).

Answer 43

-2 and 4?

Answer 44

Think of all pairs of numbers that multiply to -8.
List them and list their sum.
When you find a pair of such numbers that adds to 2, that is the pair of numbers you need to fill in the binomials above, and your factoring will be completed.

Answer 45

-9 and 1, -4 - 4, -2 - 6,

etc.

Answer 46

-2 and 4 meet the demands, correct?

Answer 47

yes correct

Answer 48

hooray!

Answer 49

@boldjon yea

Answer 50

@mathstudent55 What's next?

Answer 51

You already found that -2 * 4 = -8, and -2 + 4 = 2, so you have your numbers.

Just as a note, you could have done this:
List all pairs of numbers that multiply to -8. Follow each pair with the sum.
-8 * 1 = -8; -8 + 1 = -7
-1 * 8 = -8; -1 + 8 = 7
-2 * 4 = -8; -2 + 4 = 2 <------- this is the only pair with the product and sum we need
-4 * 2 = -8; -4 + 2 = -2

Answer 52

Now that we have our two numbers, we just fill in the blanks I left above:

\(f(t) = -16(t - 2)(t + 4) \)

Answer 53

Now we look at each factor, and we see it is no longer factorable, so the original trinomial is factored.

Answer 54

wait, where did you get the add to 2 and multiply to 8 anyways?

Answer 55

nvm found out

Answer 56

Look at the numbers in red and blue above.

Answer 57

yea i found otut

Answer 58

So the final answer is f(x) = 16(t−2)(t+4)?

Answer 59

*-16

Answer 60

Yes, with the -16.

Answer 61

Wait, hold on.

Basically I'm supposed to be finding the zeroe's and so I had to factor to do that.

I'm supposed to find the x-intercept, where do I go from here?

Answer 62

General Rule

To factor a quadratic trinomial of the type (notice the coefficient of 1 for the \(x^2\) term):

\(x^2+ax+b\)

Find two numbers (let's call them \(p\) and \(q\))
whose product is \(b\) and whose sum is \(a\).

Then the factoring is

\((x+p)(x+q)\)

Answer 63

??

Answer 64

Ok.
To find zeros, you need to set the function equal to zero.

\(f(t) = -16t^2 - 32t + 128 = 0\)

\(-16t^2 - 32t + 128 = 0\)

\(-16(t - 2)(t + 4) = 0\)

So far we have this with the factored polynomial.

Answer 65

right

Answer 66

Now we solve for t.

\(-16(t - 2)(t + 4) = 0\)

Answer 67

We have a product of factors equaling zero.
A product can only equal zero if at least one of the factors equals zero.
This is the zero product rule.

Answer 68

\(-16(t - 2)(t + 4) = 0\)

Since -16 is obviously not equal to zero, then the only factors that may equal zero are
t - 2 and t + 4.

Answer 69

alright...

Answer 70

therefore, t = 2 and t = -4

Answer 71

We set the two equations:

\(t - 2 = 0\) or \(t + 4 = 0\)

Can you solve those equations?

Answer 72

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Answer 73

Oh, you did already.

Answer 74

The zeros of the polynomial are t = 2 and t = -4.
You are correct.

Answer 75

but what happens to the -16?

Answer 76

The zeros of a polynomial are the points where the polynomial crosses the x-axis.

Answer 77

so where does teh -16 go?

Answer 78

IBANGUOAENGPOAUIEGNEAOSIGNAOEGNOPAEGOAWIEGNAEOMF

^Tantrum

Answer 79

The -16:
We had
\(-16(t - 2)(t +4)= 0\)

Divide both sides by -16:

\(\dfrac{\cancel{-16}(t - 2)(t + 4)}{\cancel{-16}~~1} = \dfrac{0}{-16}\)

The -16's cancel out on the left side.
On the right side, 0/(-16) is simply 0.

We end up with only

\((t - 2)(t + 4) = 0\)

Which has roots (or zeros) 2 and -4.
The -16 does not gives us any more zeros.

Answer 80

I don't really get it... but okaay

Answer 81

anyways, thanks for helping!

Answer 82

The -16 was simply canceled out by dividing both sides by -16.

Answer 83

Can you give a medal to @marihelenh , she/he helped too.

Answer 84

Sure.

Answer 85

The zeros are the x-intercepts.

Answer 86

You're welcome.

Answer 87

@mathstudent55 Can I tag you in another question?