= means divide. There are 2 ways to perform this division. Long Division or Synthetic Division. means divide x 2 into 3x 3 + x 2 4x 13
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1 Section 4 1B: Synthetic Division means divide There are 2 ways to perform this division. Long Division or Synthetic Division Long division requires you divide the binomial into the polynomial. means divide into Synthetic Division is a shortcut method for long division. It requires you divide the the root of the binomial into the polynomial. If the denominator is x a divide by a. If the denominator is x + a divide by a Example 1! Example 2 5x 3 + 3x 2 + x + 7 You could do "long division" ) or you could use use synthetic division and divide by 2 You could do "long division" ) x + 7 5x 3 + 3x 2 + or you could use use synthetic division and divide by Section 4 1B Lecture! Page 1 of 8! 2018 Eitel
2 Synthetic Division Use Synthetic Division to divide Example 1 Step 1: If the denomitor is divide by 2 Write the 2 and put the lines to the right of it and under it Write the coefficient of each variable term but do not write the variable Leave some space below these numbers and draw a line under the coefficients Step 2: Bring the first coefficient (the 3) stright down below the line 3 Step 3: Mutilply the 2 in the box by the 3 to get 6. Put the 6 directly under the second coefficient (the 1) and above the line 6 3 Step 4: Combine the 1 and 6 to get 7. Put the 7 below the line directly under the 1 and Step 5: Mutilply the 2 in the box by the 7 to get 14. Put the 14 directly under third coefficient (the 4) and above the line Section 4 1B Lecture! Page 2 of 8! 2018 Eitel
3 Step 6: Combine the 4 and 14 to get 10. Put the 10 below the line directly under the 4 and Step 7: Mutilply the 2 in the box by the 10 to get 20. Put the 20 directly under forth coefficient (the 13) and above the line Step 7: Combine the 13 and 20 to get 7. Put the 7 below the line directly under the 13 and x 2 x # Remainder The original numerator in the problem had the highest powered term as x 3. The polynomial answer will start with a next lowest power x 2 and then have an x term next and then a constant term. The coefficients of the first 2 terms are the first 2 numbers in the bottom row. The next number in the bottom row is the constant term in the answer. The last number in the bottom row is the remainder. We write the remainder as a fraction over the divisor Answer: 3x 2 + 7x NOTE: The example above rewrites each step with a new division graph. When you do the problem all the steps are done on the same division graph. The entire problem with all the steps included will look like the division below Single Row Method Answer: 3x 2 + 7x Section 4 1B Lecture! Page 3 of 8! 2018 Eitel
4 ! Example 2 Use Synthetic Division to divide x 3 + 2x 2 5x 10 Step 1: If the denomitor is divide by 3 Write the 3 and put the lines to the right of it and under it. Write the coefficient of each variable term but do not write the variable. Leave some space below these numbers and draw a line under the coefficients Step 2: Bring the first coefficient (the 1) stright down below the line Step 3: Mutilply the 3 in the box by the 1 to get 3. Put the 3 directly under the second coefficient (the 2) and above the line 3 1 Step 4: Combine the 2 and 3 to get 1. Put the 1 below the line directly under the 2 and Step 5: Mutilply the 3 in the box by the 1 to get 3. Put the 3 directly under third coefficient (the 5) and above the line Section 4 1B Lecture! Page 4 of 8! 2018 Eitel
5 Step 6: Combine the 5 and 3 to get 2. Put the 2 below the line directly under the 5 and Step 7: Mutilply the 3 in the box by the 2 to get 6. Put the 6 directly under forth coefficient (the 10) and above the line Step 7: Combine the 10 and 6 to get 4. Put the 4 below the line directly under the 10 and x 2 x # remainder Answer: 1x The original numerator in the problem had the highest powered term as x 3. The polynomial answer will start with a next lowest power x 2 and then have an x term and then a constant term. The coefficients of the 2 terms are the first 2 numbers in the bottom row. The next number in the bottom row is the constant term in the answer. The last number in the bottom row is the remainder. We write the remainder as a fraction over the divisor Single Row Method Answer: 1x Section 4 1B Lecture! Page 5 of 8! 2018 Eitel
6 Use Synthetic Division to find the answer to the following problems: Example 1! Example 2! Example x 2 + 7x ! x 3 + 2x 2 5x x 2! x 3 + 6x 2 + 5x 12 x x 2 + 2x 3 Note: In examples 2 and 3 the remainder is zero. This means that the denominator divides into the numerator evenly. When this happens the binomial in the denominator is a factor of the original polynomial. Synthetic Division with higher ordered polynomials. The original numerator in the problems below have the highest powered term as x 4. The polynomial answer will start with the next lowest power x 3 and then have an x 2 and then an x term and then a constant term. The coefficients of the 3 terms are the first 3 numbers in the bottom row. The next number in the bottom row is the constant term in the answer. The last number in the bottom row is the remainder. We write the remainder as a fraction over the divisor.! Example 4! Example 5 x 4 + 5x 3 + 8x 2 + 5x x 4 9x 3 5x 2 + 3x 15 x ! x 3 + 2x 2 + 2x 1+ 7! 2x 3 + x Section 4 1B Lecture! Page 6 of 8! 2018 Eitel
7 Some polynomials have missing powers of x. Each of the missing powers has a ZERO coefficient. A zero must be included in the terms when you write the polynomial under the division sign.! Example 6! Example 7!! 4x 3 2 x +1 4x 3 2 is really 4x 3 + 0x x 2 4x + 2! 3x x + 2 3x is really 3x 3 + 0x 2 + 0x x 2 6x x + 2! Section 4 1B Lecture! Page 7 of 8! 2018 Eitel
8 The Remainder Theorem When a Polynomial P(x) is divided by 1x a using long division the reminder of the division is P(a) Note: This theorem can be used only when the coefficient of the x term is 1 To find P(x) Using Synthetic Division to find P(x) 1. Use Synthetic Division to divide by x (no sign change). 2. The remainder of the Synthetic Division will be P(x) Use the Synthetic Division to find P(c) for the given P(x)! Example 2! Example 2 P(x) 4x 4 5x 3 2x 2 3x 4 find P(2)! P(x) 5x 4 + 9x 3 + 2x 2 + 3x + 7 find P( 1) the remainder is 5 so P(2) 6! the remainder is 0 so P( 1) 0 Check:! Check: P(2) 4(2) 4 5(2) 3 2(2) 2 3(2) 4 4(16) 5(8) 2(4) 3(2) P( 1) 5( 1) 4 + 9( 1) 3 + 2( 1) 2 + 3( 1) + 5 5(1) + 9( 1) + 2(1) + 3( 1) ! 0 Section 4 1B Lecture! Page 8 of 8! 2018 Eitel
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