Plowing Through Sec. 2.4b with Two New Topics: Synthetic Division Rational Zeros Theorem

Synthetic Division Synthetic Division is a shortcut method for the division of a polynomial by a linear divisor, x – k. Notes: This technique works only when dividing by a linear polynomial… It is essentially a “collapsed” version of the long division we practiced last class…

Synthetic Division – Examples: Evaluate the quotient: 32–3–5– Coefficients of dividend: Zero of divisor: Remainder Quotient

Synthetic Division – Examples: Divide by and write a summary statement in fraction form. –210–23–3 1–22–1–1 –24–42 Verify Graphically?

Rational Zeros Theorem Real zeros of polynomial functions are either rational zeros or irrational zeros. Examples: The function has rational zeros –3/2 and 3/2 The function has irrational zeros – 2 and 2

Rational Zeros Theorem Suppose f is a polynomial function of degree n > 1 of the form with every coefficient an integer and. If x = p /q is a rational zero of f, where p and q have no common integer factors other than 1, then p is an integer factor of the constant coefficient, and q is an integer factor of the leading coefficient.

RZT – Examples: Find the rational zeros of The leading and constant coefficients are both 1!!!  The only possible rational zeros are 1 and –1…check them out: So f has no rational zeros!!! (verify graphically?)

RZT – Examples: Find the rational zeros of Potential Rational Zeros: Factors of –2 Factors of 3 Graph the function to narrow the search… Good candidates: 1, – 2, possibly –1/3 or –2/3 Begin checking these zeros, using synthetic division…

RZT – Examples: Find the rational zeros of 134–5– Because the remainder is zero, x – 1 is a factor of f(x)!!! Now, factor the remaining quadratic… The rational zeros are 1, –1/3, and –2

RZT – Examples: Find the polynomial function with leading coefficient 2 that has degree 3, with –1, 3, and –5 as zeros. First, write the polynomial in factored form: Then expand into standard form:

RZT – Examples: Using only algebraic methods, find the cubic function with the given table of values. Check with a calculator graph. x–2–115 f(x) (x + 2), (x – 1), and (x – 5) must be factors… But we also have :

A New Use for Synthetic Division in Sec. 2.4: Upper and Lower Bounds

What are they??? A number k is an upper bound for the real zeros of f if f (x) is never zero when x is greater than k. A number k is a lower bound for the real zeros of f if f (x) is never zero when x is less than k.

Let’s see them graphically: c d c is a lower bound and d is an upper bound for the real zeros of f

Upper and Lower Bound Tests for Real Zeros Let f be a polynomial function of degree n > 1 with a positive leading coefficient. Suppose f (x) is divided by x – k using synthetic division. If k > 0 and every number in the last line is nonnegative (positive or zero), then k is an upper bound for the real zeros of f. If k < 0 and the numbers in the last line are alternately nonnegative and nonpositive, then k is a lower bound for the real zeros of f.

Cool Practice Problems!!! Prove that all of the real zeros of the given function must lie in the interval [–2, 5]. The function has a positive leading coefficient, so we employ our new test with –2 and 5: 52–7– –22–7–8148 –422–2828 2–1114–1436 This last line is all positive!!! This last line has alternating signs!!!  5 is an upper bound  –2 is a lower bound Let’s check these results graphically…

Cool Practice Problems!!! Find all of the real zeros of the given function. From the last example, we know that all of the rational zeros must lie on the interval [–2, 5]. Next, use the Rational Zero Theorem…potential rational zeros: Factors of 8 Factors of 2 Look at the graph to find likely candidates:Let’s try 4 and –1/2

Cool Practice Problems!!! Find all of the real zeros of the given function. 42–7– –16–8 21–4–20 –1/221–4–2 –102 20–40

Cool Practice Problems!!! Find all of the real zeros of the given function. The zeros of f are the rational numbers 4 and –1/2 and the irrational numbers are – 2 and 2

Cool Practice Problems!!! Prove that all of the real zeros of the given function lie in the interval [0, 1], and find them. Check our potential bounds: 01000– –31  0 is a lower bound!!! – –  1 is an upper bound!!! –6 8 2

Cool Practice Problems!!! Prove that all of the real zeros of the given function lie in the interval [0, 1], and find them. Possible rational zeros: Check the graph (with 0 < x < 1) to select likely candidates… The function has no rational zeros on the interval!!! Lone Real Zero: Are there any zeros???