1 / 10

Factoring Polynomials of Higher Degree

Factoring Polynomials of Higher Degree. Factoring Polynomials of Higher Degree. To review: What is the remainder when you divide x 3 – 4x 2 – 7x + 10 by x – 2?. Quotient. Divisor. Dividend. Remainder. Factoring Polynomials of Higher Degree. Remainder Theorem:

georgette
Télécharger la présentation

Factoring Polynomials of Higher Degree

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Factoring Polynomials of Higher Degree

  2. Factoring Polynomials of Higher Degree • To review: What is the remainder when you divide x3 – 4x2 – 7x + 10 by x – 2? Quotient Divisor Dividend Remainder

  3. Factoring Polynomials of Higher Degree Remainder Theorem: If a polynomial p(x) is divided by the binomial x – a, the remainder obtained is p(a). So, looking at our example, if p(x) = x3 – 4x2 – 7x + 10 was divided by x – 2, the remainder can be determined by finding p(2). p(x) = x3 – 4x2 – 7x + 10 p(2) = (2)3 – 4(2)2 – 7(2) + 10 = 8 – 16 – 14 + 10 = -12

  4. Factoring Polynomials of Higher Degree The Factor Theorem (Part 1) If p(a) = 0, then x – a is a factor of p(x) So let’s see what happens if p(1). p(1) = (1)3 – 4(1)2 – 7(1) + 10 = 1 – 4 – 7 + 10 = 0  Since p(1) = 0, we know that x – 1 is also a factor of p(x).

  5. Factoring Polynomials of Higher Degree Example 1: Determine the remainder when p(x) = x3 + 5x2 – 9x – 6 is divided by x – 3. Solution: When p(x) is divided by x – 3, the remainder is p(3). p(3) = 33 + 5(3)2 – 9(3) – 6 = 27 + 45 – 37 – 6 = 39 Thus, the remainder is 39. Since the remainder is not 0, x – 3 is not a factor of p(x).

  6. Factoring Polynomials of Higher Degree • In order to assist with the factoring of higher order polynomials, we use a process known as synthetic division. • A quick process to divide polynomials by binomials of the form x – a and bx – a.

  7. Factoring Polynomials of Higher Degree • Looking at our example: • Synthetic Division • Create a “L” shape • Place “a” toward the upper left of the “L” • Record the coefficients of the polynomials inside the “L” 2 1 -4 -7 10 Write in decreasing degree and 0’s need to be recorded as coefficients for any missing terms.

  8. Factoring Polynomials of Higher Degree Synthetic Division 4) Bring down 1 5) Multiply 1 by 2 and record result under – 4 6) Add -4 to 2 and record answer below the “L” 7) Repeat process until you reach the last term 8) The number on the furthest right below the “L” is the remainder. 2 1 -4 -7 10 2 -4 -22 1 -2 -11 -12 Remember that 1 is the number of x2’s, -2 is the number of x’s, and -11 is the constant term of the quotient polynomial.

  9. Factoring Polynomials of Higher Degree Example 2: Use synthetic division to find the quotient and remainder when x3 – 4x2 – 7x + 10 is divided by x – 5. 5 1 -4 -7 10 5 5 -10 Remainder 0 means that x – 5 is a factor of p(x) 1 1 -2 0

  10. Homework • Do # 1, 3, 5, 7, 9, 17, 18, 21 on pg 130 in Section 4.3 for Monday  Have a great weekend!!!

More Related