Answer:
![(x-43.45)(x+7.45)=0](https://tex.z-dn.net/?f=%28x-43.45%29%28x%2B7.45%29%3D0)
Step-by-step explanation:
We have the quadratic equation
i.e. ![d^{2}-36d-324=0](https://tex.z-dn.net/?f=d%5E%7B2%7D-36d-324%3D0)
As, the roots of the quadratic equation
are given by
.
So, from the given equation, we have,
a = 1, b = -36 , c = -324.
Substituting the values in
, we get,
![x=\frac{36\pm \sqrt{(-36)^{2}-4\times 1\times (-324)}}{2\times 1}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B36%5Cpm%20%5Csqrt%7B%28-36%29%5E%7B2%7D-4%5Ctimes%201%5Ctimes%20%28-324%29%7D%7D%7B2%5Ctimes%201%7D)
i.e. ![x=\frac{36\pm \sqrt{1296+1296}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B36%5Cpm%20%5Csqrt%7B1296%2B1296%7D%7D%7B2%7D)
i.e. ![x=\frac{36\pm \sqrt{2592}}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B36%5Cpm%20%5Csqrt%7B2592%7D%7D%7B2%7D)
i.e. ![x=\frac{36\pm 50.9}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B36%5Cpm%2050.9%7D%7B2%7D)
i.e.
and ![x=\frac{36-50.9}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B36-50.9%7D%7B2%7D)
i.e.
and ![x=\frac{-14.9}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-14.9%7D%7B2%7D)
i.e. x = 43.45 and x = -7.45
Thus, the roots of the equation are 43.45 and -7.45.
Hence, the factored form of the given expression will be ![(x-43.45)(x+7.45)=0](https://tex.z-dn.net/?f=%28x-43.45%29%28x%2B7.45%29%3D0)
I'm going to assume that you are asking how a number can have two factors.
So, a factor is an integer (non-fractional/decimal number) you multiply by another integer to get the actual number.
Well, let's look at the number 4. The number four has the following factors:
1, 2, 4
In a 'rainbow' style, you match the two outer numbers, then the two next inner numbers, and so on.
Because the number '2' is the one in the center and there is no other number to multiply it by, it must multiply by itself to reach the number 4.
Well, you must have two numbers to actually multiply together, so every number has two factors: 1, and itself (except for 0).
I hope this helps, and good luck :)
Answer:
2x(x - 4)
Step-by-step explanation:
![2 {x}^{2} - 8x \\ \\ = 2x(x - 4)](https://tex.z-dn.net/?f=2%20%7Bx%7D%5E%7B2%7D%20%20-%208x%20%5C%5C%20%20%5C%5C%20%20%3D%202x%28x%20-%204%29)
Let us denote the smaller even integer by
![N](https://tex.z-dn.net/?f=N)
.
Then, the next even integer must be
![N+2](https://tex.z-dn.net/?f=N%2B2)
.
To verify, set
![N](https://tex.z-dn.net/?f=N)
as any even number, say
![174](https://tex.z-dn.net/?f=174)
.
Then, the next even integer must be 2 more than this number, i.e.
![174+2=176](https://tex.z-dn.net/?f=174%2B2%3D176)
.
The sum of these two consecutive even integers can be expressed as:
![N+(N+2)=2N+2](https://tex.z-dn.net/?f=N%2B%28N%2B2%29%3D2N%2B2)
We are told that this sum divided by four is 189.5, i.e.
![(2N+2)\div4=189.5](https://tex.z-dn.net/?f=%282N%2B2%29%5Cdiv4%3D189.5)
Multiplying both sides by 4 will reverse the division:
![(2N+2)\times1=758](https://tex.z-dn.net/?f=%282N%2B2%29%5Ctimes1%3D758)
Subtracting 2 from each side will reverse the addition:
![2N=756](https://tex.z-dn.net/?f=2N%3D756)
Dividing both sides by 2 will reverse the multiplication:
![N=378](https://tex.z-dn.net/?f=N%3D378)
Therefore, the two consecutive integers must be
378 and
380.
Check this:
6 is your answer, hope it helped