Solving Higher Order Polynomials

Solving Higher Order Polynomials

We will be able to...

~Use synthetic division to solve polynomials

~Factor to solve polynomials

~Determine functions and factors from graphs

Open Ended

Given the graph, state all the x values where the function intersects the x axis. (You can click on the graph to make it bigger.)

Multiple Choice

The graph has a zero at x=-7. The factor that goes with this zero is...?

• (x+7)

• (x-7)

• (x+0)

Multiple Choice

The graph has another zero at x=2. The factor that goes with this zero is...?

• (x+2)

• (x-2)

• (x+0)

Multiple Select

There are two more zeros, choose the factors that go with the zeros that are left over.

• (x+1)

• (x-1)

• (x+9)

• (x-9)

Building a Function

~We can put all these factors together to write a function that could represent the graph.

~We also need to consider the a value!

Multiple Choice

Considering all the factors and the end behavior, what is a function that could match the graph?

• $f\left(x\right)=8\left(x+1\right)\left(x+7\right)\left(x-2\right)\left(x-9\right)$

• $f\left(x\right)=\left(x-1\right)\left(x-7\right)\left(x+2\right)\left(x+9\right)$

• $f\left(x\right)=-2\left(x+1\right)\left(x+7\right)\left(x-2\right)\left(x-9\right)$

Open Ended

Analyze the new graph.  Considering all the the zeros, the factors that go with them and the end behavior, write a function that could possibly represent the graph.

Open Ended

Analyze the new graph.  Considering all the the zeros, the factors that go with them and the end behavior, write a function that could possibly represent the graph.

Open Ended

Analyze the new graph.  Considering all the the zeros, the factors that go with them and the end behavior, write a function that could possibly represent the graph.

Solve:

 $0=2x^3+3x^2-18x-27$  

• Graph, determine as many zeros as possible from graph

• Synthetically divide until you run out of zeros

• Solve with factoring, quadratic formula, or inverse operations

Multiple Select

Solve the problem with me, then select ALL the solutions to let me know you're there!

 $0=2x^3+3x^2-18x-27$  

• x=2

• x=-3

• x=3

• x=-2/3

• x=-3/2

Multiple Select

Graph the function to find whole number zeros, do not include decimals.

 $0=2x^3-x^2-27x+36$  

• x=3

• x=-2

• x=1

• x=-4

Multiple Choice

You should have seen that the graph crosses at x=-4 and 3.  We can take one of those zeros and divide.  Doesn't matter which one... Let's divide by 3 first.

 $0=2x^3-x^2-27x+36$  

•  $2x^2+5x+12$  
  

•  $2x^2+5x-12$  
  

•  $2x^3+5x^2+12x$  
  

•  $3x^2+8x-3$  
  

Open Ended

Write a function in factored form that COULD represent the graph based on it's zeros and end behavior.

Multiple Choice

Solve: First graph. What is the only zero we can see?

 $x^3-x^2+49x-49=0$  

• x=2

• x=-2

• x=1

• x=-1

Multiple Choice

Solve: Since x=1 is the only zero, we need to divide by it.  What is the result?

 $x^3-x^2+49x-49=0$  

•  $x^3+49x$  
  

•  $x^3+49x^2$  
  

•  $x^2+49x$  
  

•  $x^2+49$  
  

Multiple Choice

Solve: We are left a quadratic we can solve with inverse operations. What is the first step to solve for x?

 $x^2+49=0$  

• take the square root of both sides

• add 49

• subtract 49

• divide by 49

Multiple Choice

Solve: Now we can take the square root of both sides.  What would x equal?

 $x^2=-49$  

• 7

•  $\pm7i$  
  

•  $\sqrt{7}$  

•  $\pm7$  
  

Multiple Choice

Solve: First graph. What is the only zero we can see?

 $x^3-7x^2+37x+45=0$  

• x=2

• x=-2

• x=1

• x=-1

Multiple Choice

Solve: Synthetically divide by the zero x=-1. What is the result?

 $x^3-7x^2+37x+45=0$  

•  $x^2-8x+45$  
  

•  $x^2-6x+43$  
  

•  $x^2-8x+29$  
  

•  $x^2+8x+29$  
  

Multiple Choice

Solve: We can't solve this with inverse operations, we have to factor or use...dun dun DUN... QUADRATIC FORMULA!?!?!?

 $x^2-8x+45=0$  

•  $\frac{-8\pm\sqrt{\left(-8\right)^2+4\left(1\right)\left(45\right)}}{2\left(1\right)}$  
  

•  $\frac{8\pm\sqrt{\left(-8\right)^2-4\left(1\right)\left(45\right)}}{2\left(1\right)}$  
  

•  $\frac{8\pm\sqrt{\left(-8\right)+4\left(1\right)\left(45\right)}}{2\left(1\right)}$  

Multiple Choice

Solve: When we do the quadratic formula, what do you get as the discriminate?  (Number under the square root.)

 $\frac{8\pm\sqrt{\left(-8\right)^2-4\left(1\right)\left(45\right)}}{2\left(1\right)}$  

• -172

• -224

• -116

• -360

Multiple Choice

Solve: Now we have to break down the number under the square root! Remember!?! What perfect square can we divide -116 by?

 $\frac{8\pm\sqrt{-116}}{2\left(1\right)}$  

• 4

• 9

• 16

• 25

Multiple Choice

Solve: We can use 4 and 29 to break down the radical.  What would our answer look like after taking the square roots?

 $\frac{8\pm\sqrt{-1}\sqrt{4}\sqrt{29}}{2}$  

•  $\frac{8\pm4i\sqrt{29}}{2}$  
  

•  $\frac{8\pm2i\sqrt{29}}{2}$  

•  $\frac{8\pm2\sqrt{29}}{2}$  
  

Multiple Choice

Solve: Last step!  Reduce the numerator by the denominator if you can. 

 $\frac{8\pm2i\sqrt{29}}{2}$  

•  $4\pm i\sqrt{29}$  
  

•  $\frac{4\pm i\sqrt{29}}{2}$  

•  $8\pm i\sqrt{29}$  
  

•  $4\pm2i\sqrt{29}$  

That was ALOT of steps... but it's not so bad.

• Graph to find whole number zeros

• Continually synthetically divide until you run our of zeros

• Solve by factoring, quadratic formula, or inverse operations

Multiple Choice

Solve: First graph. What is the only zero we can see?

 $x^3-2x+4=0$  

• x=2

• x=-2

• x=1

• x=-1

Multiple Choice

Solve: x=-2 is the only zero on the graph so divide by it.  What do you get?

 $x^3-2x+4=0$  

•  $x^2-4$  
  

•  $x^2-4x+12$  
  

•  $x^2-2x+2$  
  

•  $x^3+2x^2-6x+16$  
  

Multiple Choice

Solve: Now we decide how to solve.  Factor if you can, otherwise we need to do quadratic formula.

 $x^2-2x+2=0$  

•  $\frac{2\pm\sqrt{\left(-2\right)^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}$  
  

•  $\frac{-2\sqrt{\left(-2\right)^2+4\left(1\right)\left(2\right)}}{2\left(1\right)}$  
  

•  $\frac{-2\pm\sqrt{\left(-4\right)^2-4\left(1\right)\left(-2\right)}}{2\left(1\right)}$  
  

Multiple Choice

Solve: Determine the discriminant

 $\frac{2\pm\sqrt{\left(-2\right)^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}$  

•  $\frac{2\pm\sqrt{-4}}{2\left(1\right)}$  
  

•  $\frac{2\pm\sqrt{4}}{2\left(1\right)}$  
  

•  $\frac{2\pm\sqrt{-32}}{2\left(1\right)}$  
  

•  $\frac{2\pm\sqrt{32}}{2\left(1\right)}$  

Multiple Choice

Solve: Break down the radical.

 $\frac{2\pm\sqrt{-4}}{2\left(1\right)}$  

•  $\frac{2\pm2}{2\left(1\right)}$  
  

•  $\frac{2\pm2i}{2\left(1\right)}$  
  

•  $\frac{2\pm4i}{2\left(1\right)}$  
  

•  $\frac{2\pm4}{2\left(1\right)}$  

Multiple Choice

Solve: Reduce if you can

 $\frac{2\pm2i}{2\left(1\right)}$  

•  $\pm i$  
  

•  $2\pm2i$  
  

•  $1\pm i$  
  

•  $2\pm i$  

Multiple Choice

Solve: x=-3 is the only zero on the graph so divide by it.  What do you get?

 $2x^3+9x^2+14x+15=0$  

•  $2x^2+3x+5$  

•  $2x^2+15x+59+\frac{162}{x+3}$  

•  $2x^3+3x^2+5x$  

•   $x^2+3x-5$  
  

Multiple Choice

Solve: Now we decide how to solve.  Factor if you can, otherwise we need to do quadratic formula.

 $2x^2+3x+5=0$  

•  $\frac{-3\pm\sqrt{\left(3\right)^2-4\left(2\right)\left(5\right)}}{2\left(2\right)}$  
  

•  $\frac{3\pm\sqrt{\left(3\right)^2-4\left(2\right)\left(5\right)}}{2\left(1\right)}$  
  

•  $\frac{-3\sqrt{\left(3\right)^2+4\left(2\right)\left(5\right)}}{2\left(2\right)}$  
  

Multiple Choice

Solve: Determine the discriminant

 $\frac{-3\pm\sqrt{\left(3\right)^2-4\left(2\right)\left(5\right)}}{2\left(2\right)}$  

•  $\frac{-3\pm\sqrt{-31}}{2\left(2\right)}$  
  

•  $\frac{-3\pm\sqrt{-49}}{2\left(2\right)}$  
  

•  $\frac{-3\pm\sqrt{-37}}{2\left(2\right)}$  
  

•  $\frac{-3\pm\sqrt{49}}{2\left(2\right)}$  

Multiple Choice

Solve: Break down the radical.

 $\frac{-3\pm\sqrt{-31}}{2\left(2\right)}$  

•  $\frac{-3\pm i\sqrt{31}}{4}$  
  

•  $\frac{-3\pm31i}{4}$  
  

•  $\frac{-3\pm31}{4}$  
  

•  $\frac{-3\pm\sqrt{31}}{4}$  

Multiple Select

Solve: First graph. What TWO zeros do you see on the graph?

 $x^4-3x^3+12x-16=0$  

• x=-2

• x=2

• x=4

• x=-4

• x=0

Multiple Choice

Solve: Since there are two zeros, you will divide twice in any order.  Just be sure to continue dividing, don't divide the original problem twice

 $x^4-3x^3+12x-16=0$  

•  $x^2-3x+4$  

•  $x^3-3x^2+4x$  

•  $3x^2+2x-5$  

•   $x^2-x+4$  
  

Multiple Choice

Solve: Now we decide how to solve.  Factor if you can, otherwise we need to do quadratic formula.

 $x^2-3x+4=0$  

•  $\frac{3\pm\sqrt{\left(-3\right)^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}$  
  

•  $\frac{3\sqrt{\left(-3\right)^2+4\left(1\right)\left(4\right)}}{2\left(1\right)}$  
  

•  $\frac{-3\pm\sqrt{\left(-3\right)^2+4\left(1\right)\left(4\right)}}{4\left(1\right)}$  
  

Multiple Choice

Solve: Now simplify as much as possible

 $\frac{3\pm\sqrt{\left(-3\right)^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}$  

•  $\frac{3\pm i\sqrt{7}}{2}$  
  

•  $\frac{3\pm\sqrt{7}}{2}$  
  

•  $\frac{3\pm7i}{2}$