Warm up!!
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Transcript of Warm up!!
1) Divide 2) Factor 3) Find Zeros
Example:Divide 6x3 – 19x2 + 16x – 4 by x – 2
Answer: 6x2 – 7x + 2Which means that…
6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2)6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2)
Zeros: x = 2 x = ½ x = ⅔
Answer: 5x + 3
Can check zeros on
calculator
Rules of Long Division
If the divisor goes into the equation evenly then it is a factor and a zero.
If the quotient has a remainder then it is not a factor.The proper form of writing the remainder is adding it
to the quotient and over the divisor.
Example:Divide x3-1 by x-1
– Write in descending powers– Insert zeros where there are missing power
Because it equals 0, x3 -1 is divisible by x-1
x2 + x + 1x – 1 √(x3 + 0x2 + 0x – 1)
x3 – x2
x2 + 0x x2 – x
x – 1 x – 1
0
Using Division Algorithm you can write it like this:
x3 – 1 = x2 + x + 1 x – 1
• Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor. – Used when finding zeroes of polynomials.
Example: Divide x2 + 5x + 6 by x – 1
*Rational roots test determines that +/- 1, 2, 3, and 6 are possible zeros.
• Synthetic provides the same quotient but in a quicker fashion.
• If r = 0, (x – k) is a factor.• If r = 0, (k, 0) is an x intercept of f. *k is a zero• The remainder r gives the value of f at x = k, if
r = f(k) (Remainder Theorem)
To evaluate polynomial function f(x) when x=k, divide f(x) by x-k
– Remainder will equal f(k)
Example 1:Evaluate f(x) = 3x3 + 8x2 + 5x – 7 for x = -2
So f(-2) = -9, r = -9
Example 2:Evaluate f(x) = x3 – x2 – 14x + 11 for x = 4 So f(4) = 3, r = 3
Repeated Division Multi-step synthetic division…
Find all the zeros of the following polynomial function given (x – 2) and (x + 3) are factors…
f (x) = 2x4 + 7x3 – 4x2 – 27x – 18.
Solution:Using synthetic division with the factor (x – 2), you obtain the following.
0 remainder, so f (2) = 0 and (x – 2) is a factor.
Repeated Division (Continued)Take the result of this division and perform synthetic division again using the
factor (x + 3).
Because the resulting quadratic expression factors as2x2 + 5x + 3 = (2x + 3)(x + 1)
the complete factorization of f (x) is f (x) = (x – 2)(x + 3)(2x + 3)(x + 1).
0 remainder, so f (-3) = 0 and (x +3 ) is a factor.
Try #60 on page 141
Example 1:Show that (x – 2) and (x + 3) are factors and find all the zeros off(x) = 2x4 + 7x3 – 4x2 – 27x – 18
x = 2, -3, -3/2, -1
Example 2:Show that (x + 2) and (x – 1) are factors and find all the zeros off(x) = 2x3 + x2 – 5x + 2
x = ½, -2, 1
• Graph should confirm zeros