So you have your polynomial. To factor this out, you need to find the factors of -45 that will add up to 3. These two factors would be -5 and 8. So, you have factored your polynomial! (x-5)(x+8). If you are still unsure, simply use the FOIL method to check and see if you get the same polynomial back. Hope this helps!
Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
X +x²= 132
x +x²-132= 132-132
x²+x -132 = 0
(x-11) (x+12) = 0
x-11 = 0 X+12=0
x-11+11 = 0+11 X+12-12=0-12
x =11 x= -12
Check
x +x²= 132 x +x²= 132
-12 +-12²= 132 11 +11²= 132
-12 + 144 = 132 11+ 121=132
132= 132 132=132
Since the problem says positive the answer is (11,121)
You used the fact that you can multiply both sides of an equality by the same, non-zero constant.
So, if you start with the trivial equality
you can multiply both sides by 3:

But if
,
becomes
, which is
. So, we have

which leads to

if you recall that 