3+2 x20
3+2=5
5x20=
100
Hope this helped.
difference of squares means that both the terms are square terms. (also there must be a - symbol)
for example
y^2 - 4
square root of y^2 is y
square root of 4 is +2 as well as -2
so you would factorise it like this:
(y+2)(y-2)
1. y^4 has a square root of y^2 as y^2 × y^2 is y^4.
<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>-2 doesnt have a whole number square root so it is not a difference of squares.
2. 25 has a square root of 5. m^2 has a square root of m. n^4 has a square root of n^2. so this 25m^2n^4 is a square term.
1 has a square root of +1 and -1.
therefore, this one is a difference of squares. <u>(</u><u>5</u><u>m</u><u>n</u><u>^</u><u>2</u><u> </u><u>+</u><u>1</u><u>)</u><u> </u><u>(</u><u>5</u><u>mn^2</u><u> </u><u>-</u><u>1</u><u>)</u>
3. p^8 has a square root of p^4. q^4 has a square root of +q^2 and -q^2)
so it is a difference of squares. <u>(</u><u>p</u><u>^</u><u>4</u><u>+</u><u>q</u><u>^</u><u>2</u><u>)</u><u>(</u><u>p</u><u>^</u><u>4</u><u> </u><u>-</u><u>q</u><u>^</u><u>2</u><u>)</u>
4. 16x^2 is a square term as irs square root is 4x.
<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>24 is not a square term.
therefore, it is not a difference of squares.
In this problem, we need to plug in the given x values for
![f(x)=0](https://tex.z-dn.net/?f=%20f%28x%29%3D0%20)
and find a and b.
When we plug in 1, we get:
![2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0](https://tex.z-dn.net/?f=2%20%5Ctimes%20%7B1%7D%5E%7B4%7D%20-%205%20%5Ctimes%20%7B1%7D%5E%7B3%7D%20-%2014%20%5Ctimes%20%7B1%7D%5E%7B2%7D%20%2B%20a%20%5Ctimes%201%20%2B%20b%20%3D%200)
Simplify:
![2 - 5 - 14 + a + b = 0](https://tex.z-dn.net/?f=2%20-%205%20-%2014%20%2B%20a%20%2B%20b%20%3D%200)
![- 17 + a + b = 0](https://tex.z-dn.net/?f=%20-%2017%20%2B%20a%20%2B%20b%20%3D%200)
![a + b = 17](https://tex.z-dn.net/?f=a%20%2B%20b%20%3D%2017)
We got our first statement about the values of the variables. If we find one more we can find those 2 variables.
We have another given root: 4.
Plug it in:
![2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0](https://tex.z-dn.net/?f=2%20%5Ctimes%20%7B4%7D%5E%7B4%7D%20-%205%20%5Ctimes%20%7B4%7D%5E%7B3%7D%20-%2014%20%5Ctimes%20%7B4%7D%5E%7B2%7D%20%2B%20a%20%5Ctimes%204%20%2B%20b%20%3D%200)
![512 - 320 - 224 + 4a + b = 0](https://tex.z-dn.net/?f=512%20-%20320%20-%20224%20%2B%204a%20%2B%20b%20%3D%200)
![- 32 + 4a + b = 0](https://tex.z-dn.net/?f=%20-%2032%20%2B%204a%20%2B%20b%20%3D%200)
![4a + b = 32](https://tex.z-dn.net/?f=4a%20%2B%20b%20%3D%2032)
Now we have our second one. We can combine them:
![a + b = 17 \\ 4a + b = 32](https://tex.z-dn.net/?f=a%20%2B%20b%20%3D%2017%20%5C%5C%204a%20%2B%20b%20%3D%2032)
I use elimination method which is easier here.
Multiply the top equation by -1:
![- a - b = - 17 \\ 4a + b = 32](https://tex.z-dn.net/?f=%20-%20a%20-%20b%20%3D%20-%2017%20%5C%5C%204a%20%2B%20b%20%3D%2032)
Add them up:
![3a = 15](https://tex.z-dn.net/?f=3a%20%3D%2015)
Simplify:
![a = 5](https://tex.z-dn.net/?f=a%20%3D%205)
Now we have a, we can plug in one of those equations to find b:
![5 + b = 17](https://tex.z-dn.net/?f=5%20%2B%20b%20%3D%2017)
![b = 17 - 5](https://tex.z-dn.net/?f=b%20%3D%2017%20-%205)
![b = 12](https://tex.z-dn.net/?f=b%20%3D%2012)
So, the answers are
![a=5](https://tex.z-dn.net/?f=a%3D5)
and
![b=12](https://tex.z-dn.net/?f=b%3D12)
.