Chapter 4

Polynomial Long Division Remember that time you did long division in elementary school and thought you’d never see it again?

Polynomial Long Division In higher level math, we use long division with polynomials to help us break down the function from standard to factored form.

Let’s take it back…way back. Back to elementary school kind of back.

Now we take it a step further…

Practice Makes Perfect…right?

Warm Up Page 15 in ACT Prep Book Answer both questions

Synthetic Division Yay! Another way to divide that’s basically a short hand for long division. ONLY works when dividing by a linear factor

Synthetic Division First, write down the coefficients in descending order. The k (from the factor x-k) goes on the outside. Be sure to flip the sign! Bring down the 1 st coefficient and multiply by k Add the column Repeat the process The coefficient of the remainder is the last column Write the answer using variables starting with the exponent that is one less than the dividend’s degree

When there’s a coefficient before x in x – k… Take the k and divide it by the coefficient That’s your factor For example, if you’re given 2x – 3, the value you divide by is 3/2. Try: 3x + 1 8x – 2 2x + 4 9x - 2

Warm Up Open ACT book to page 13 Do #1-5 When completed, do page 14 #1-5

f(x) = 3x 2 + 8x - 4 Divide by x – 2 (using synthetic or long division) Find f(2) –Substitute 2 into your equation for x

g(x) = x 3 - 3x 2 + 6x + 8 Divide by x + 1 Find g(-1)

h(x) = x 3 + 2x - 3 Divide by x – 3 Find h(3)

Looking at the results of the quotient and the substitution, what pattern do we see?

Remainder Theorem The remainder theorem states that if a polynomial f(x) is divided by a linear divisor x – a, the remainder is f(a).

Use synthetic division with the given value to find your remainder

Factor Theorem If you divide a polynomial by a factor x – a and there is a zero remainder, then x – a is a factor for that polynomial. In other words, if it divides evenly, it’s a factor. Thinking about long division, is 8 a factor if 14?

Using Factor Theorem to Find Unknown Coefficients Given the question: Determine the value of a if x + 1 is a factor of f(x)= x 3 + 4x 2 – ax + 1 Any ideas on how we could answer this?

Substitute the factor into the equation Remember to flip the sign! Set your answer equal to zero Solve for a

Determine a if x – 8 is a factor of f(x) = x 4 + ax 3 – 5x 2 – 21x – 24

Determine a if x – 2 is a factor of f(x) = 2x 3 – x 2 – ax + 10

Determine a if x – 3 is a factor of f(x) = 3x 3 – ax x + 3

Warm Up

What is factoring? We have been faced with ALL kinds of functions Factoring gives us a method to break those functions down into their parts Just like how standard form and factored form were the same in quadratic functions

Snowflake Factoring of Quadratic Functions x 2 + 5x + 6 2x 2 + 8x + 6 2x 2 + x - 3 3x 2 + 7x + 2

Snowflake Factoring 7x 2 – 45x – 28 2x x x 2 – x – 18 7x 2 – 32x – 60

Factoring using GCF Determine the greatest common factor of ALL THREE terms Factor using the snowflake method

GCF

Grouping

Factoring the Sum or Difference of Cubes ONLY when you have pairs of cubes

Difference of Squares ONLY when you have a pair of squares being subtracted

Perfect Square Trinomial