Unit 3 Test Prep

Unit 3 Test Prep

by Morgan Hale

Factors and Equations​

Multiple Choice

If I were to tell you that (x + 2) is a factor of  $$x^6+2x^5+4x^4+16x^3+16x^2+32x+64$$  , the first thing you should do is....?

• Try and factor by grouping

• Divide

• Graph

Multiple Choice

Same as the last slide, but now I want you to actually divide   $$\left(x^6+2x^5+4x^4+16x^3+16x^2+32x+64\right)\div\left(x+2\right)$$  

• $$x^2+4x+8+\frac{32}{x+2}$$  

• $$x^4+4x^2+8x+\frac{32}{x+2}$$  

• $$x^3+4x^2+8x+32$$  

• $$x^5+4x^3+8x^2+32$$  

Multiple Choice

Using your answer from the last slide where we divided $$\left(x^6+2x^5+4x^4+16x^3+16x^2+32x+64\right)\div\left(x+2\right)$$  ,

I can factor

$$\left(x^6+2x^5+4x^4+16x^3+16x^2+32x+64\right)$$   like .....

• $$\left(x+2\right)\left(x^5+4x^3+8x^2+32\right)$$  

• $$\frac{\left(x^5+4x^3+8x^2+32\right)}{\left(x+2\right)}$$  

• You can't factor the polynomial because there are too many terms.

Open Ended

Can I continue to factor the following factorization?

$$\left(x+2\right)\left(x^5+4x^3+8x^2+32\right)$$  

If so, which method would I start with?

Multiple Choice

If I were to factor the quintic factor from the last slide using grouping, 

$$x^5+4x^3+8x^2+32$$  

What would I end up with for my next step? 

• $$\left(x^3+8\right)\left(x^2+4\right)$$  

• $$\left(x^2+8\right)\left(x^3+4\right)$$  

• $$\left(x^2-4\right)\left(x^3+8\right)$$  

• $$\left(x^2-4\right)\left(x^3-8\right)$$  

Open Ended

Can I factor $$\left(x^3+8\right)\left(x^2+4\right)$$   any further?

If yes, using which method?

​The answer is yes, we can factor it further using the sum of cubes formula.

Solving an Equation ​

Open Ended

Now that I know I can factor the polynomial 

$$x^6+2x^5+4x^4+16x^3+16x^2+32x+64$$  

into

  $$\left(x+2\right)^2\left(x^2+4\right)\left(x^2-2x+4\right)$$  

How can I solve the equation 

$$x^6+2x^5+4x^4+16x^3+16x^2+32x+64=0$$  ?