Notes for Algebra 2 Honors 6



Notes for Algebra 2 Honors 6.5 Continued and 6.6

Since not every polynomial is easily factored and some may be prime, the Possible Rational Roots Theorem can be used to create a list of POSSIBLE RATIONAL ROOTS.

To do this, you look at the factors of the lead coefficient and the factors of the constant. The possible rational roots, PRR, is the quotient of [pic].

Identify the PRR of P(x)

Example 1A: P(x) = 2x3 - 5x2 - 4x + 10

Factors of Constant 1, 2, 5, 10

Factors of Lead Coefficient 1, 2

Notice that this list provides some duplicates. You only need to list items once. Don’t forget the ± symbol though.

Finding the roots of P(x)

Now that the list has been narrowed down from all real numbers to just the list of PRR, synthetic substitution can be used to identify roots or solutions of the equation.

To do this, select any number from the PRR and apply the Remainder Theorem.

Use your calculator to help you make a good guess!

Example 1B: Select [pic]

Since the remainder is -2, we know that 2 is not a root. Cross it off your list. Don’t eliminate -2 though until you are sure it is not a root. Continue selecting values to use in the synthetic substitution. STOP whenever you identify a root.

Now let’s pick [pic] .

Since the remainder is 0, then we know that [pic] is a root. We also know that when P(x) is divided by [pic], that the quotient is 2x2 +0x -4. This is called the depressed equation. We can continue solving by using the depressed polynomial.

If the depressed polynomial is higher than 2nd degree, you should refine your list of PRR using the constant and lead coefficient of the depressed polynomial and continue attempting to find roots. REMEMBER USE YOUR CALCULATOR TO HELP YOU FIND ROOTS so that you don’t make bad guesses!

If the depressed polynomial is second degree, then solve by any of the methods we have discussed for solving quadratics.

Example 1C: 2x2 - 4 =0

x2 - 2 =0

x2 = 2

x = ±[pic]

Therefore, the real roots of P(x) = 2x3 - 5x2 - 4x + 10 are x = [pic].

Sometimes you may have imaginary roots involved. Sometimes you may need to use the quadratic formula to solve the quadratic depressed polynomial. For instance, you might have a solution of [pic] as a root. If the depressed polynomial is x2+5, then the solutions would be ±[pic].

EXAMPLE 2:

Find all roots of P(x) = x4 -3x3 + 5x2 - 27x - 36

Factors of Constant 1, 2, 3, 4, 6, 9, 12, 18, 36

Factors of Lead Coefficient 1

Try 6

We can now cross +6 off the list.

Try 4

This means that x = 4 is a root. The depressed polynomial is x3 +x2 +9x +9.

The depressed polynomial has PRR of ±1, ±3, ±9. IF IT HAD CONTAINED 6, THEN WE COULD HAVE IMMEDIATELY TAKEN 6 off the list since it had already been eliminated. Now repeat the process of synthetic substitution with this new polynomial and the new list of PRRs.

Try -1

The depressed polynomial is now x2 +9.

This can be solved by x2+9 =0

x2 = -9

x = [pic].

The roots of the polynomial equation are: x=-1, x = 4, and x = [pic] .

Writing the simplest polynomial function with the given zeros.

Example 3A: A polynomial has zeros (roots) of -4, 5, 2. Find the polynomial in simplest form. This is where you need to think backwards.

We know that x = -4, x = 5, and x = 2.

Therefore, x + 4 = 0 x -5 = 0 and x -2 = 0.

That means that the polynomial had factors of (x+4), (x-5), and (x-2).

Therefore, f(x) = (x+4) (x-5)(x-2).

Now multiply it out to find the polynomial in simplest form.

f(x) = (x2-x-20)(x-2)

f(x) = x3-x2-20x -2x2 +2x +40

f(x) = x3-3x2 -18x +40 is the polynomial in simplest form.

Using the calculator to approximate solutions of an equation by finding all roots.

Just as we have used the calculator to find the roots of a quadratic by using the Calculate Zeros (found under 2nd Calculate), we can also use this same process with polynomial functions of degree greater than 2.

Example 4A: [pic]

Enter this into the y= screen of your calculator. Be sure to close the parentheses after the 7. The x3 is not part of the radicand.

Y1 = [pic] x ^ 3 – 11 x 2 + 8

Now use 2nd calc to find the zeros.

When asked for left and right bounds, be careful to

place the cursor to the left & to the right of the root

you are trying to locate. When asked to make a guess,

just hit enter. The arrows on the right indicate

the location of the cursor whenever I selected those

boundaries.

Round to the nearest thousandth.

x= -.782 x= .975 x=3.965

-----------------------

PRR = [pic]

2

2 -5 -4 10

4 -2 -12

2 -1 -6 -2

[pic]

2 -5 -4 10

5 0 -10

2 0 -4 0

[pic]

PRR: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

-6

6

1 -3 5 -27 -36

6 18 138 666

1 3 23 111 630

4

1 -3 5 -27 -36

4 4 36 36

1 1 9 9 0

-1

1 1 9 9

-1 0 -9

1 0 9 0

This indicates that x = -1 is a root.

[pic]

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