You can form a perfect square trinomial from x² + bx by adding (b/2)²

x² + bx + (b/2)²

Quadratic Formula

-b±√b²-4ac

2a

b² - 4ac

Value>0 two solutions

Value=0 one solution

Value<0 no solutions

Multiplying and Dividing Complex Numbers

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To DIVIDE, multiply conjugates

Finding Rational Roots

2x³ - x² + 2x + 5

 factor of constant term

factor of leading coefficient

Constant Factors(P):±1,±5

Leading Factors(Q): ±1,±5,±½,±5/2

P/Q, then test the roots

If a + √b is a root, then so is a - √b

If a + bi is a root, then so is a - bi

• Use Descartes' Rule of Signs to determine the number of real zeroes of:
  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5.

Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect.

First, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case:

f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5

Ignoring the actual values of the coefficients, I then look at the signs on those coefficients:

f (x) = +x⁵ – x⁴ + 3x³ + 9x² – x + 5

I draw little lines underneath to highlight where the signs change from positive to negative or from negative to positive from one term to the next:

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Then I count the number of changes:

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There are four sign changes in the "positive" case. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5. However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complexand thus not graphable as x-intercepts. Because of this possibility, I have to count down by two's to find the complete list of the possible number of zeroes. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).

Now I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case):

f (–x) = (–x)⁵ – (–x)⁴ + 3(–x)³ + 9(–x)² – (–x) + 5

= –x⁵ – x⁴ – 3x³ + 9x² + x + 5

I look at the signs:

f (–x) = –x⁵ – x⁴ – 3x³ + 9x² + x + 5

...and I count the number of sign changes:

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There is only one sign change in this "negative" case, so there is exactly one negative root. (In this case, I don't try to count down by two's, because the first subtraction would give me a negative number.)

There are 4, 2, or 0 positive roots, and exactly 1 negative root.

Rhythm of Solving Polynomials

1. Write in Standard form, highest degree indicates # of zeros
2. List possible zeros P/Q
3. Use Remainder Theorem to find a factor that gives a remainder of zero, or use Synthetic Division
4. Once you have successfully divided enough times to get a quadratic equation, factor and find the remaining zeros
5. If you can't factor, use the quadratic formula

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