Factor completely
x³-8x²+11x+20
To factor the polynomial x³-8x²+11x+20 completely, we can use a combination of methods including grouping and factoring by grouping.
First, we look for common factors among the terms. In this case, there are no common factors.
Next, we can look for any factor pairs of the constant term (20) and the coefficient of the leading term (1). The potential factor pairs are (1,20), (2,10), and (4,5).
By trying each factor pair in turn, we can determine if any of them will yield a factor of the polynomial when tested using synthetic division or substitution.
Let's try the first factor pair (1,20):
Using synthetic division, we divide the polynomial x³-8x²+11x+20 by (x-1):
```
1 | 1 -8 11 20
|___________
1 -7 4 24
```
The result of the synthetic division is x²-7x+4 with a remainder of 24.
Since the remainder is not zero, (x-1) is not a factor. Moving on to the next factor pair.
Now, let's try the factor pair (2,10):
Using synthetic division, we divide the polynomial x³-8x²+11x+20 by (x-2):
```
2 | 1 -8 11 20
|___________
2 -12 -2
2 -4 7
```
The result of the synthetic division is x²-4x+7 with a remainder of 2.
Again, the remainder is not zero, which means (x-2) is not a factor. Let's try the last factor pair.
Now, let's try the factor pair (4,5):
Using synthetic division, we divide the polynomial x³-8x²+11x+20 by (x-4):
```
4 | 1 -8 11 20
|___________
4 -16 -20
4 -12 -9
```
The result of the synthetic division is x²-12x-9 with a remainder of 0.
Finally, we found a factor! (x-4) is indeed a factor of the given polynomial.
To factor further, we can divide the polynomial x²-12x-9 by (x-4). By synthetic division:
```
4 | 1 -12 -9
|__________
4 -32 -44
1 -28 -53
```
The result of the synthetic division is x-28 with a remainder of -53.
This gives us the factored form:
(x³-8x²+11x+20) = (x-4)(x²-12x-9)
Therefore, the polynomial x³-8x²+11x+20 can be fully factored as (x-4)(x²-12x-9).
try some factors of 20
I tried ±1 and sure enough -1 worked, so x+1 is a factor
Then by either long division or synthetic division of x³-8x²+11x+20 by x+1
we would get x^2 - 9x + 20
which factors again into (x-4)(x-5)
so x³-8x²+11x+20 = (x+1)x-4)(x-5)