Consider the two expressions \(\frac{6x^3+9x^2+4x+7}{2x+3}\) and \(3x^2+2+\frac1{2x+3}\)

a) Show that the two expressions represent equal numbers when \(x=10\)

b) Explain why these two expressions do not represent equal numbers when \(x=-\dfrac32\)

c) Show that these two expressions represent equal numbers for all \(x\) other than \(-\dfrac32\)

In parts (a) and (c), begin by explaining what your strategy for solving will be.

Can you pleaes provide a written explanation of the proccess during the different steps? Just so I can understand better. Thanks!

onyulee Jun 25, 2024

#1**+1 **

a) Setting these two equations to equal to each other, we have

\(\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac1{2x+3}\)

Now, if we plug 10 into the equation and it returns equal values, then 10 is an answer.

However, if the two equations don't equal each other, then we know that 10 is not an answer.

Plugging in 10, we have

\(\frac{6(10^3)+9(10^2)+4(10)+7}{2(10)+3}=3(10^2)+2+\frac1{2(10)+3}\)

Simplifying both sides, we get that

\(\frac{6000+900+40+7}{20+3}=300+2+\frac{1}{20+3}\\ 6947/23 = 302 + 1/23\\ 6947/23=6947/23\)

Thus, we can easily tell that the two equations are equal.

Thus, 10 is defintely a solution!

Thanks!: )

NotThatSmart Jun 25, 2024

#2**+1 **

b) Now, we can plug in -3/2 for x into the equation \(\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac1{2x+3}\), and determine if we achieve equal factors.

If we do, then it is a solution.

If not, then it isn't a solution.

Thus, we have

\(\frac{6(-3/2)^3+9(-3/2)^2+4(-3/2)+7}{2(-3/2)+3}=3(-3/2)^2+2+\frac1{2(-3/2)+3}\)

Now, let's focus on the denominators of the equation. We have that

\(2(-3/2)+3 = -3+3=0\)

This means that the left hand side of the equation is undefined.

Also, we also have

\(2(-3/2)+3=-3+3=0\) for the denominator in the right hand side, which is also undefined.

Thus, -3/2 is not a solution since both sides become undefined when we plug it in for x.

Thanks! :)

NotThatSmart Jun 25, 2024

#4**+1 **

c) First, we set them equal to each other to see for any value x it is true.

By setting each expresstion equal to each other, we have \(\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac{1}{2x+3}\).multiplying by 2x+3 on each side gets us \(6x^3 + 9x^2+4x+7=3x^2(2x+3)+2(2x+3)+1.\)Simlifying the right hand side leaves us with \(6x^3+9x^2+4x+7=6x^3+9x^2+4x+6+1\). As you can see, we end up with\(0=0, \)witch is a true equation, and therefore any value of x works. But, a fraction can not be diuvvide by zero, and if x=-2/3, plugged into 2x+3, makes you didvide by zero. so, All values of x work except for -3/2.

sometimg5764 Jun 25, 2024