Question

20w^2+11W-3 FACTOR!

Answer 1

@shamil98

Answer 2

@Hero

Answer 3

Find two numbers, m and n such that:

\(m \times n = -60\)
\(m + n = 11\)

Answer 4

what do you mean

Answer 5

We need to figure that out in order to start factoring

Answer 6

Actually it might help if I wrote it this way:

\(m \times n = 60\)
\(m - n = 11\)

Answer 7

There exists two numbers such that when you multiply them you get 60, but when you subtract them, you get 11

Answer 8

6x10=60

Answer 9

Yeah, but we want to find two numbers that satisfy both conditions

Answer 10

12x5=60 but its not 11 correct?

Answer 11

12 - 5 = 7

so that's not what we want either. Keep trying

Answer 12

15x4=60
15-4=11

Answer 13

Yes, that's good

Answer 14

Now, in the quadratic, replace 11 with 15 - 4 like so

\(20w^2 + (15 - 4)w - 3\)

Then distribute the w:
\(20w^2 + 15w - 4w - 3\)

Answer 15

Then factor the first two terms and the last two terms:

\(5w(4w + 3) - 1(4w + 3)\)

Then factor out the remaining common term 4x + 3:
\((4w + 3)(5w - 1)\)

Answer 16

@Sephora
The rason you need two factors of -60 that add to 11 is that in order to factor a trinomial of the form
\(ax ^2 + bx + c = 0 \)
you start by multiplying together \(a\) and \(c\).

Then you need to find two factors of the product ac that add to b.

In your case, you have
\(20w^2 + 11w - 3 = 0\)

In this case, \(a = 20\) and \(c = -3\),

Then when you find ac, you get \(ac = 20(-3) = -60\)

Now you need two factors of -60 that add to b which is 11 in your case.

Answer 17

So (4w+3)(5w-1) is the correct answer

Answer 18

Yes

Answer 19

I need help with 6a^4+a^2-1

Answer 20

now this is kind of different. i need help to factor this

Answer 21

mathstudent can you help me out

Answer 22

You can use the same method.
Because you have the 4th poer here, one way of thinking of this problem is by doing a substitution.
Let x = a^2, then x^2 = a^4.

Then the problem becomes
6x^2 + x - 1

Now use the method above to factor this.
Multiply ac together, etc.

Answer 23

So it'll be -6

Answer 24

Right, ac = -6.
Now, what two numbers multiply to -6 and add to 1?

Answer 25

2 and 3

Answer 26

Actually, -2 and 3.
Now break up the midlle term into two terms using -2 + 3 = 1

Answer 27

(-6a^2-2)(a^2+3)

Answer 28

No.
We're not factoring yet into two sets of parentheses.

After you find the two numbers that multiply to ac and add to b, you break up the middle term into two parts:

6x^2 + x - 1

6x^2 + 3x - 2x + 1 (The middle term, x, is now written as 3x - 2x. We can do this
because 3x - 2x = x).

Answer 29

Now we factor by grouping. That means we factor a common factor out of the first two terms, and we factor a common factor out of the last two terms.

Answer 30

3 and 1

Answer 31

Notice that above I accidentally typed + 1 instead of -1 as the last term of the trinomial.


\(6x^2 + 3x - 2x - 1\)

\(3x(2x + 1) -1(2x + 1) \) We factor 3x out of the first two terms
and -1 out of the last two terms.

Answer 32

NOw we see that we have a common factor of 2x + 1, so we factor it out.

\( (2x + 1)(3x - 1) \)

Answer 33

We aren't done yet. Your problem had a in it, not x, and we made a substituttion, so now we need to substitute a back in and continue.

Answer 34

okay

Answer 35

Since we let x = a^2, where we see x we must insert a^2.

\( (2a^2 + 1)(3a^2 - 1) \)

Answer 36

we just change the variable x to an a
and thats it.

Answer 37

So (2a^2+1)(3a^2-1)

Answer 38

is the correct answer?

Answer 39

Now we need to see if we can factor anything else.
The first binomial is a sum and there is no common factor between 2a^2 and 1, so it is not factorable.
The second binomial is a difference (a subtraction), but 3a^2 is not a perfect square, so this is not a difference of squares, so it is also not factorable.

This means the problem is finished and fully factored.

Answer 40

Yes, this is the final and correct answer.

Answer 41

I need help with factoring v(v+5)=0

Answer 42

It's already factored.

Answer 43

it says simplify your answer. Use a comma to separate answers as needed.

Answer 44

There is no simplifying here. This is an equation that can be solved.

In the equation, you have two factors multiplied together equaling zero. Let's look at a simple case.

In the wequation
xy = 0,
what can x and y be?

The only way a product is zero is if one or more factors are zero.

The only way xy = 0 is if x = 0 or y = 0 or both x and y are equal zero.

Answer 45

Now let's look again at your equation.

v(v + 5) = 0
The only way the product v(v + 5) can equal zero is if v = 0 or if v + 5 = 0.

The way you solve an equation like this is you set each factor equal to zero and solve each equation. Then you separate the answers with the word "or."

v(v + 5) = 0
v = 0 or v + 5 = 0
v = 0 or v = -5

The solution is v = 0 or v = -5.

Now let's check our solution.

If v = 0, then the equation would be
0(0 + 5) = 0
0 * 5 = 0
0 = 0 Is a true statement, so v = 0 works.

If v = -5, then we have
-5(-5 + 5) = 0
-5(0) = 0
0 = 0 is also true, so v = -5 also works.

Our solution was correct.

The final solution is: v = 0 or v = -5.