Dividing Polynomials. First divide 3 into 6 or x into x 2 Now divide 3 into 5 or x into 11x Long Division If the divisor has more than one term, perform.

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Presentation transcript:

Dividing Polynomials

First divide 3 into 6 or x into x 2 Now divide 3 into 5 or x into 11x Long Division If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps x - 3 x 2 + 8x x 64 x 2 – 3x Now multiply by the divisor and put the answer below. Subtract (which changes the sign of each term in the polynomial) 5 11x Bring down the next number or term Multiply and put below 32 11x - 33 subtract This is the remainder Remainder added here over divisor So we found the answer to the problem x 2 + 8x – 5  x – 3 or the problem written another way:

Divide y into -2yDivide y into y 2 Let's Try Another One If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight. y + 2 y 2 + 0y + 8 y y 2 + 2y Subtract (which changes the sign of each term in the polynomial) -2y Multiply and put below - 2y - 4 subtract 12 This is the remainder Remainder added here over divisor Write out with long division including 0y for missing term Bring down the next term Multiply and put below

List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. 1 Set divisor = 0 and solve. Put answer here. x + 3 = 0 so x = - 3 Synthetic Division There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x) Bring first number down below line Multiply these and put answer above line in next column - 3 Add these up 3 Multiply these and put answer above line in next column - 9 Add these up Multiply these and put answer above line in next column Add these up This is the remainder Put variables back in (one x was divided out in process so first number is one less power than original problem). x 2 + x So the answer is:

List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms. 1 Set divisor = 0 and solve. Put answer here. x - 4 = 0 so x = 4 Let's try another Synthetic Division Bring first number down below line Multiply these and put answer above line in next column 4 Add these up 4 Multiply these and put answer above line in next column 16 Add these up Multiply these and put answer above line in next column Add these up This is the remainder Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x 3 ). x 3 + x 2 + x + So the answer is: 0 x 3 0 x Multiply these and put answer above line in next column Add these up

List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. Let's try a problem where we factor the polynomial completely given one of its factors Bring first number down below line Multiply these and put answer above line in next column - 8 Add these up 0 Multiply these and put answer above line in next column 0 Add these up Multiply these and put answer above line in next column Add these up No remainder so x + 2 IS a factor because it divided in evenly Put variables back in (one x was divided out in process so first number is one less power than original problem). x 2 + x So the answer is the divisor times the quotient: You could check this by multiplying them out and getting original polynomial