NotThatSmart

First, let's combine all like terms. 

We get \(2x^2+sx+38\)

Dividing everything by 2, we get \(x^2 + \frac{s}{2}x+19\)

Square rooting 19, we get \(\sqrt{19}\)

So, we have \(\sqrt{19}/2 = s/2\)

So \(s=\sqrt{19}\)

Thanks! :)

First, let's combine all like terms. 

We have \(3x^{2}-32x+c\) as the simplified version. 

Now, let's divide everything by 3 to make our lives idea. 

We get \(x^2-\frac{32}{3}x+c/3\). 

Now, we divide 32/3 by 2, and we get \(16/3\)

Now, \((16/3)^2 = 256/9\). 

Now, we have 

\(c/3 = 256/9\\ c=253/3\)

Thanks! :)

Alright, we just need to find the one with something in the form of (x-a)^2. 

First, we have x^2-2x+1. That can be factored to (x-1)^2. 

Well, there's our answer. The others have distinct roots. 

Thanks! :)

Ok, we have to do a number of things to set up this equation. 

First off, we know that \((a+b)^2=a^2+2ab+b^2\)

Plugging in the values we know, we get that 

\(4^2=6+2ab\\ 10=2ab\\ ab=5\)

It is crucial for us to know this information. 

Now, let's take a look at a^3+b^3. 

We know that \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3}\)

Now, take a look at the middle two terms. We can factor them to get \((a+b)^3=a^{3}+3ab(a+b)+b^{3}\)

We already know all these terms! We get

\(4^3=3(5)(4) +a^3+b^3\\ 64-60=a^3+b^3\\ 4=a^3+b^3\)

So 4 is our answer!

Thanks! :)

First, let's combine all like terms and move all terms to one side. 

We get

\(x^2 + (17 - m)x + 4 > 0 \)

Let's note that if roots are real, then the descriminant must be greater or equal to 0. 

We have

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16 \)

From here, there are 2 ways to go. First, we have

\(17 - m \geq 4 \\ 17 - 4 \geq m \\ m \leq 13 \)

and, we have

\( 17 - m \leq -4 \\ 21 \leq m \\ m \geq 21\)

This means our final answer is

\(m = [-\infty , 13]\bigcup [21, \infty]\)

Thanks! :)

Thanks! It's funny because I'm taking an intermediate algebra course, and this is what we are learning. 

Also, NTS is such a wonderful abbreviation, LOL. Sounds quite proffesional. 

Thanks! :)

Combining all like terms and moving all terms to one side, we get 

\(x^{2}+12x-4=0\)

The two roots of this are

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Squaring them we get 

\(4\left(\sqrt{10}\right)^{2}-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

\(4\cdot 10-12\sqrt{10}-6\left(2\sqrt{10}-6\right)+\left(-2\sqrt{10}-6\right)^{2}\)

This simplifies to 152, which is our answer.

Thanks! :)

First, let's subtract 10 from both sides. \(x^2-mx+14\)

The answer to this question lies in the number of possible multiples 14 has. 

For every mutliple, there are 4 possibles. 

Say the two multiples are \(a, b\)

We have \((x-a)(x-b), (x+a)(x+b), (x+a)(x-b), (x-a)(x+b)\)

For example, 2 and 7. 

\((x-7)(x+2), (x-2)(x+7), (x-2)(x-7), (x+7)(x+2)\)These all give different values for m. 

We have only 2 possible groups. 1 and 14, 2 and 7. 

This means there are 4*2=8 different values for m

Thanks! :)

So, I might have forgotten to consider than b and c could include z. I'll solve the problem like that, 

Like I stated in post 1, there are 4 roots

\(4+2i, 4-2i, -5+8i, -5-8i\)

This means we have the polynomial, 

\((z-4-2i)(z-4+2i)(z+5-8i)(z+5+8i)\)

Simplfying, we have 

\(\left(z^2-8z+20\right)\left(z^2+10z+89\right)\\z^4+2z^3+29z^2-512z+1780\)

So, in order for this to be valid, b would have to be z^3 and c would have to be \(2z^3+28z^2-512z+1780\)

So our final answer is \(z^4+2z^3+28z^2-512z+1780\)

Thanks! :)

There's a problem with this question, though. 

According some theorem I totally forgot, 

If \(4 + 2i\) is a root, then \(4-2i\) must also be a root. 

If \(-5 + 8i\) is a root, then \(-5-8i\) must also be a root.

With 4 distinct roots, then the polynomial would have degree 4, not 2, 

So it is impossible to have a quadratic with these two roots. 

I'm not sure if I'm mistaken, but that should be true. 

Thanks! :)