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(18/2)²   =   9²   =   81

81  needs to be added to both sides to complete the square on the left side.

3.

(a² + 3a - 7)(2a² - a + 4)

=  a²(2a²) + a²(-a) + a²(4) + 3a(2a²) + 3a(-a) + 3a(4) + (-7)(2a²) + (-7)(-a) + (-7)(4)

=  2a⁴  +  -a³  +  4a²  +  6a³  +  -3a²  +  12a  +  -14a²  +  7a  +  -28

=  2a⁴  +  5a³  -  13a²  +  19a  -  28

y  =  2x^2 - 4x + 4    and    y  =  -x^2 - 2x + 4

First let's find the  x  values of the intersection points by solving this equation:

2x^2 - 4x + 4   =   -x^2 - 2x + 4

                                                  Add  x^2  to both sides of the equation.

3x^2 - 4x + 4   =   -2x + 4

                                                  Add  2x  to both sides.

3x^2 - 2x + 4   =   4

                                                  Subtract  4  from both sides.

3x^2 - 2x  =  0

                                                  Factor  x  out of the two terms on the left side.

x(3x - 2)   =   0

                                                  Set each factor equal to zero.

x  =  0     or     3x - 2  =  0

                       3x  =  2

                         x  =  2/3

The  x  coordinates of the intersection points are  0  and  2/3 .

So   c = 2/3   and   a = 0

c - a   =   2/3 - 0   =   2/3

Consecutive angles in a parallelogram are supplemental....so....

B + C  = 180

110  +  C   = 180      subtract 110 from both sides

C  = 70°

For factoring, I generally use the "AC-method." The method does not have a formal name, so I will assign it myself! In a quadratic, which is in the form \(\textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}\), you first consider factors of \(\textcolor{red}{a}\textcolor{green}{c}\) that sum to \(\textcolor{blue}{b}\).

In this particular case for \(\textcolor{red}{5}x^2+\textcolor{blue}{22}x+\textcolor{green}{8}\),  \(\textcolor{red}{a}\textcolor{green}{c}=\textcolor{red}{5}*\textcolor{green}{8}=40\) and \(\textcolor{blue}{b}=\textcolor{blue}{22}\). We need to find two numbers that multiply to get 40 and add to get 22. Let's list a few factors.

Of the factors of 40 listed, only 2 and 20 are factors that multiply to 40 and add to 22. We can use this information to manipulate the given trinomial.

Factoring is one of those computations where there are a multitude of ways to arrive at the same answer. This is just the method I generally default to--even if it is not necessarily most efficient. 

\(49x^8-16y^{14}\) can be written as a difference of squares.

Solve.

What is the solution of the equation when solved over the complex numbers?

Solve for x over the complex numbers.

What is the factored form of the expression over the complex numbers?

New Year, New Start, New Answer!

First answer of the year for me.

Alright, I'll try to explain this the best I can using a keyboard. 

The equation is x^2(6x - 11)(7x - 8) = 0

So, you have to figure out what you can put in for x for the whole thing to equal 0.

Roots mean possible solutions.

Let's "divide" the equation into 3 parts. 

We know that anything multiplied by 0 is 0.

And 0 to the power of anything is still 0. 

So if you plug in 0 for x^2(6x - 11)(7x - 8) = 0, You get 0 (it would look like "0(6x-11)(7x-8)" )

So the first root is 0.

Next, let's look at the (6x - 11) part. 

Remember that anything multiplied by 0 is 0. 

So, 6x - 11 must equal to 0.

6x-11=0

6x=11

x=11/6

So another root is 11/6.

Finally, let's look at (7x-8) 

That should also equal 0, no matter what it is multiplied by.

7x-8=0

7x=8

x=8/7

The last root is 8/7

So the three roots are 0, 11/6, and 8/7.

I beg to differ just a little EP

By convention the answer must be positive and we do not know if a or b is pos or neg.

\(\sqrt[4]{256a^2b^4}\\ =\sqrt[2]{16|a|b^2}\\ =4\sqrt{|a|}|b|\\ =|\;4b\sqrt{|a|}\;|\\ or\\ =|\;4b|a|^{1/2}\;|\\\)

\(p^2+4p=1\\ p^2+4p+(\frac{4}{2})^2=1+(\frac{4}{2})^2\\ p^2+4p+4=1+4\\ (p+2)^2=5\\ p+2=\pm\sqrt5\\ p=-2\pm\sqrt5\\\)

Q1

\(3x^3+4x^2-15x-20\\ x^2(3x+4)-5(3x+4)\\ (x^2-5)(3x+4)\)

(7x+3)(4x-5)

You take the first term in the first bracket and multiply it by the second bracket

so that is 7x     times by    (4x+5)

then you take the second term    +3   and multiply it by the second bracket.

If there were more terms would keep doing this but there are no more terms here.

So we have

7x(4x-5)      +3(4x-5)

now you just have to multiply out these brackets and then collect up the like terms. 

You do it and I will check you working :)

That is pretty neat!

Thanks for sharing :)

On the graph of y = (x + 2)^4 - 100, how many points are there whose coordinates are both negative integers?

When is y negative?

\((x+2)^4-100<0\\ (x+2)^4<100\\ -\sqrt{10}

That is 5 points where the coordinates are both negative integers:

(-5, -19),  (-4,-84),   (-3,-99),   (-2,-100),  (-1,-99)

I just answered this here:

I consulted with a higher power and got a great answer!  Thank you  

Suppose f(x)  is a polynomial of degree 4 or greater such that f(1)=2,   f(2)=3, and    f(3)=5. Find the remainder when f(x)    is divided by    (x-1)(x-2)(x-3).

\(\begin{align} \frac{f(x)}{(x-1)(x-2)(x-3)}&=g(x)+\frac{ax^2+bx+c}{(x-1)(x-2)(x-3)}\\ \frac{f(x) \times (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}&=g(x)(x-1)(x-2)(x-3)+\frac{ax^2+bx+c}{(x-1)(x-2)(x-3)}\times (x-1)(x-2)(x-3)\\ f(x)&=g(x)(x-1)(x-2)(x-3)+ax^2+bx+c\\ now\;\;substitute\\ f(1)&=a+b+c=2\\ f(2)&=4a+2b+c=3\\ f(3)&=9a+3b+c=5\\ \end{align} \)

So now I have 3 equations and 3 unknowns so I just have to solve them simultaneously.

I solved them using matrices.  These are the unknowns.

\(c=2, \;\; b=\frac{-1}{2} \;\;and\;\;a=\frac{1}{2}\\ \text{So when f(x) is divided by (x-1)(x-2)(x-3) the remainder is } \frac{x^2}{2}-\frac{x}{2}+2\)