00:01
There are a lot of parts to this problem, but for each one, you're going to go through the same steps.
00:07
You're going to start by guessing and checking to try and find a factor.
00:12
A factor is one that when you plug it in for x, you get zero.
00:17
Once you find a factor, you're going to do synthetic division so that you're left with a linear term times a quadratic.
00:26
And then the third step will be to factor the quadratic if possible.
00:31
So we're going to go ahead and jump right in and start by checking x equal to 1.
00:37
So if x is equal to 1, then this becomes 1 plus 6 plus 5 minus 12, which is equal to 0.
00:48
So we got lucky on our first try.
00:51
We found a factor.
00:52
And so we're going to set up synthetic division with our box.
00:56
So our factor 1 and our box here where we put the terms, or sorry, the coefficients from the polynomial.
01:05
So 1, 6, 5, and negative 12.
01:11
The 1 drops down.
01:14
Now we multiply 1 times 1 is 1.
01:18
Add 6 plus 1 is 7.
01:21
Multiply 7 times 1 is 7.
01:25
7 plus 5 is 12.
01:28
12 times 1 is 12 and 12 plus negative 12 is 0.
01:34
Perfect.
01:35
So now we're left with x minus 1 because our factor was 1 and then x squared plus 7x plus 12.
01:51
Now we need to factor this quadratic.
01:55
My favorite way of factoring is to find the multiple, or sorry, not the multiples, the factors of the last term, and then see which two of them combine to add to the middle term.
02:07
So the factors of 12 are 1 and 12, 2 and 6, and 3 and 4.
02:16
So when we're thinking about what could get plugged in here, we want those two factors to add to 7, 3 plus 4 adds to 7.
02:29
So we know one of these must be x plus.
02:32
3 and the other must be x plus 4.
02:37
So that's part a.
02:38
I'm going to erase to move on to the next one, but you want to follow the same steps.
02:44
So i'll leave the steps up here.
02:46
We'll again start by guessing and checking until we find a factor that works.
02:51
Then we'll use synthetic division, and then finally factor the quadratic.
02:58
So the next problem is 2x cubed.
03:03
Plus 9x squared minus 11x minus 30.
03:09
And if there's a coefficient in front of the leading term, you first want to make sure that you can't factor it out.
03:18
In this case, we can't because it's not true that all of the coefficients are multiples of two.
03:27
And so we will just go ahead and move on to checking to find a factor.
03:35
So let's start again with x equals 1.
03:39
In that case we end up with 2 plus 9 minus 11 minus 30 and that's going to end up being equal to negative 30.
03:50
So x plus 1 is not a factor in this case or sorry x equal to 1.
03:57
Now we will try negative 1.
04:00
So that makes this a negative 2 plus 9 plus 11 minus 30 and this doesn't work either because that all sums to negative 12.
04:16
So now i'm going to increase and look at x equal to 2 and see if that will work.
04:22
So 2 times x cubed or sorry 2 times 2 cubed or 2 times 8 is 16 for the first term plus 9 times 2 squared or 9 times 4 gives us 36 for the second term.
04:43
Minus 11 times 2 is minus 22 for the third term and then minus 30.
04:49
And this does all sum to 0, so we finally found our factor and we can move on to the second step synthetic division.
04:58
So again, we'll set up our box with the coefficients to 9, negative 11, and negative.
05:07
30, bring down the two, multiply, and add, multiply to get 26, and add to get 15, 15 times 2 is 30, and we add and get 0.
05:26
So great, we found our factor.
05:30
So that leaves this as being x minus 2 times 2x squared plus 13x plus 15.
05:44
And so again we need to think about factoring the quadratic.
05:49
I will again look at the factors of 15.
05:52
So that's 1 and 15 and 3 and 5.
05:56
But it's a little bit more complicated this time because one of those factors.
06:03
Is going to be multiplied by two.
06:05
To get a leading term of 2x squared, we're going to have a 2x and an x as the leading terms in our two linear formulas, or equations here.
06:20
Okay, so now we want to figure out how can we get to 13, add up to 13.
06:27
And so i clearly see that if you multiply 5 times 2, you get 10 and then add to 3 gets 13.
06:36
So that means i want my 5 to be here so that when i multiply the outside terms when foiling, i get 10, and then 3 here to get 3.
06:46
And when you add, you get 15.
06:49
Okay, we've tackled part b, but there are a lot more to go.
06:55
So again, i'm going to erase to make space for the next one.
07:00
But we're again going to follow that same process.
07:04
All of these follow the same process of guessing and checking to find a factor, doing synthetic division, and then factoring the quadratic.
07:13
So i'll probably go a little bit faster through the rest of these because we now have done this process a few times and sort of know what to expect.
07:25
So for part c, we have three x cubed.
07:30
Minus 4 x squared, minus 28x minus 16.
07:38
I can try x equals 1.
07:41
That would be 3 minus 4, minus 28, minus 16, which i can already tell is not going to be 0, so that will not be a factor.
07:54
We can try negative 1, and we would end up with negative 3 minus 4.
08:01
Plus 28 minus 16.
08:05
When i add that up, i get 5.
08:08
So still not a factor.
08:12
We can.
08:13
I'm not even going to try out positive 2 because it's, again, going to be way too negative like positive 1 was, and we'll skip right to negative 2.
08:27
So negative 2 gives me negative 24 for the first term.
08:31
We get minus 16 plus 56 minus 16.
08:38
And that equals to zero.
08:41
So we've found our factor.
08:42
We can move to step two, synthetic division.
08:47
So again, we set up our box and put the terms, or coefficients, rather.
08:56
Bring down the three.
08:57
Multiply, add, multiply, add, multiply, add, we get zero, which means we were right.
09:13
And that leaves us with x plus two times three x squared minus 10x minus eight.
09:25
Again, we're going to look at the factors of eight we get one and eight and two and four.
09:32
These have to subtract to add to 10 when one of them is multiplied by three.
09:39
Again, because the leading coefficient here is three.
09:44
And i know that they have to subtract because there's a negative here.
09:52
So when i'm thinking about that, i think about the fact that 4 times 3 is 12 and 12 minus 2 is 10.
10:03
So that is exactly what we want.
10:07
And so we're left with x plus 2.
10:11
We'll have 3x and x.
10:14
We just stated we want the x to be multiplied by or sorry the 4 to be multiplied by 3.
10:20
So it goes there.
10:21
The 2 goes here.
10:22
And our middle term is negative.
10:24
So we need the larger thing to be negative, which means we're subtracted.
10:29
For and adding to so that we end up when we're foiling with negative 12x plus 2x, which would be negative 10x.
10:43
Chugging right along to part d.
10:50
Again, we'll use those same steps of guessing and checking, synthetic division, and factoring the quadratic for part d.
11:07
3 x cubed minus x squared plus x plus 5.
11:16
Again, i'm going to skip some steps here.
11:21
You would continue to guess and check, but i already know that x equal to negative 1 is the factor because i would get negative 3 minus 1, minus 1 plus 5 equal to 0.
11:34
So we can jump into synthetic division, bring down the three, multiply, add, multiply, add, multiply, and add, which leaves us at this point with x plus 1 and 3x squared minus 4x plus 5.
12:06
And so our factors here are 1 and 5.
12:10
We need those factors to multiply to 5, but add to 4 when one of them is multiplied by 3.
12:23
And it turns out that this is impossible because we could multiply the 1 by 3 and we would have 3 plus 5, but that's equal to 8, not 4.
12:35
Or we could multiply the 5 by 3 and get 1 plus 15, which is 16, which is also not equal to 4.
12:44
So this is the end of the road here.
12:47
We're not able to further factor the quadratic in this case.
12:54
Moving on to the next one.
13:00
So here we start with 6x cubed minus 5x squared, minus 5x squared, minus 34x plus 40.
13:15
Again, i'm going to skip through some of my guesses because i know that they're wrong, but eventually if you were doing this, you would get to x equals 2, and you would see that 48 minus 20, minus 68 plus 40 is equal to zero.
13:33
So that's your factor.
13:35
You would set up synthetic division, with a coefficients in the box, bring down the six, and again, multiply and add, multiply and add, multiply and add.
13:59
That leaves you in this case with x minus two times six x squared plus seven x minus 20.
14:12
And so it's a little bit more complicated this time because we've got this coefficient of 6 here in front of the leading term.
14:24
And this is negative, which tells us that our two things must subtract to equal 7.
14:32
So we can go ahead and find the factors of 20.
14:35
We have 1 and 20, 2 and 10, and four and five.
14:44
But we additionally have to think about that six.
14:47
So the factors of six are going to be one and six or two and three...