Help with this solution. Question 22 from the latest algebra 2 common core June regents

Question

which value to the nearest tenth is not a solution of p(x)=q(x) if p(x)=x^3+3x^2-3x-1 and q(x)=3x+8

 

thank you!

Answer 1

p(x) = q(x)

 

x³ + 3x² - 3x - 1 = 3x + 8

 

 

Move all the terms to the left side to make the right side equal to zero.

 

x³ + 3x² - 6x - 9 = 0

 

 

Next, we can use synthetic division, to find a linear factor and quadratic factor that can multiplied together to give us a product of the left side.  Or we can try to factor by group factoring.  Or plot the function

 

y = x³ + 3x² - 6x - 9

 

on a graphing calculator, and find the zeros from there.  Since this is a cubic equation, you will have at most 3 distinct solutions.

 

Synthetic division and group factoring did not work here, so use the graphing calculator if you have one.

 

 

 

Answer 2

p(x)=x^3+3x^2-3x-1 and q(x)=3x+8  means 

 

x^3+3x^2-3x-1 =3x+8, i.e.

 

x³ + 3x² - 6x - 9 = 0.  

 

You can test each proferred answer (none of which you have provided) to see which one would work.

 

 

Answer 3

Where are the values?

 

p(x) = x³ + 3x² - 3x -1 ;   q(x) = 3x + 8 

 

Setting p(x) = q(x)

 

            x³ + 3x² - 3x -1 = 3x + 8

 

           X³ + 3X² - 6x -9 = 0

 

You could use synthetic division to find the zeros of the function, 

these would be your points of intersection of p(x) and q(x)

 

Do you use a Graphing Calculator?

Answer 4

first make an equation from the information given to you.

 

px)=q(x)

 

so x^3+3x^2-3x-1 = 3x + 8

 

now get everything to one side

move the 3x over and the 8

 

now you have x^3 +3x^2-6x-9=0

 

graph that on your calculator

you'll see it crosses at

-3.9, -1.1 and 2.1

 

so that leaves #4 as the one that is not a solution