Questions 3
Answers 956


B; The solutions are 5, -2, 3i, and -3i


Let's look at the original equation:




Our goal is to get the left hand side into parts that are more manageable to work with. I'll try to explain this. We'll use the rational root theorem. This says that if a polynomial equation can be written in the form \(a_nx^n+a_{n-1}x^{n-1}+...+a_0\) and if \(a_0\) and \(a_n\) are integers, then the factor can be found by checking numbers produced by doing \(\pm\frac{dividers\hspace{1mm}of\hspace{1mm}a_0}{dividers\hspace{1mm}of\hspace{1mm}a_n}\).


Dividers of \(a_0:\hspace{1mm}1,2,3,5,6,9,10,15,18,30,45,90\)  

Dividers of \(a_n:\hspace{1mm}1\)


Therefore, check for one rational solution by doing: \( \pm\frac{1,2,3,5,6,9,10,15,18,30,45,90}{1}\)


-2 happens to be the first value that satisfies this equation. The opposite of that is 2, so a factor is x+2.


\(\frac{x^4-3x^3-x^2-27x-90}{x+2}=x^3-5x^2+9x-45\), so the monstrosity we had before goes to \((x+2)(x^3-5x+9x-45)=0\).


Thankfully, this is the hardest part. Now, we have to factor \((x^3-5x^2+9x-45)\) further. Find another factor by grouping:




Factor out an x^2 from the first set of parentheses and 9 from the second set and then factor the resulting common term, which happens to be (x-5):




After all this effort, we have transformed \(x^4-3x^3-x^2-27x-90\) to \((x+2)(x-5)(x^2+9)\).


Okay, now use the zero factor principle. In other words, set each factor to zero:




Let's focus on the easy ones first:


\(x=-2 \) and \(x=5\)


Now the harder factor:







Solution set is 5,2,3i, -3i

May 21, 2017

You need not graph these equations to solve them (but the directions tell you to do so). In case you are in the situation where you are, this is what you can do:





If I were you, I would solve this by substitution. In other words, since y=-x+5, you can substitute that into the y in the other equation, so you would get. Then, simplify:






To make things simpler, factor out a 2 because that is the GCF of the left side and then factor (if possible):


\(2(x^2-5x+6)=0\)      x^2-5x+6 happens to be factorable

\(2(x-3)(x-2)=0\)     Set each factor set to 0 and solve.

\(x-3=0 \hspace{1cm} x-2=0\)



Now, plug each x-value to solve for the y-values. I'll pick the second equation as that one has simpler calculations overall:





Therefore, your solution set is:


\((2,3)\) and \((3,2)\)


Let's do that for the other problems for extra practice.


Let's list the original equations for problem 52:





Substitute x+8 into the first equation and simplify:




\(x^2+2x+1+x^2+14x+49=18\)     Combine like terms

\(2x^2+16x+32=0\)     Factor out a GCF of 2 to simplify things



Now, I could factor x^2+8x+16, but this trinomial is special: it is a perfect square trinomial. I know this because I can work backwards. The x^2 term and 16 term are both perfect squares. The square root of x^2 equals x, and the square root of 16 equals 4. 4x*2=8x. If all three conditions are met (like it is here), then it is a perfect square trinomial:






In this case, x only has one answer. This is actually easier than the previous problem because we only have to find one y-value instead of two. Now, plug in the x-value into the second equation:






Therefore, the solution set is:



YOu can apply this method for the next one, too...



May 21, 2017