 # Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor The quotient will be here. 2x³ + 3x² - x + 1 is the dividend.

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Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor The quotient will be here. 2x³ + 3x² - x + 1 is the dividend

Algebraic long division First divide the first term of the dividend, 2x³, by x (the first term of the divisor). This gives 2x². This will be the first term of the quotient.

Algebraic long division Now multiply 2x² by x + 2 and subtract

Algebraic long division Bring down the next term, -x.-x.

Algebraic long division Now divide –x², the first term of –x² - x, by x, the first term of the divisor which gives –x.–x.

Algebraic long division Multiply –x by x + 2 and subtract

Algebraic long division Bring down the next term, 1

Algebraic long division Divide x, the first term of x + 1, by x,x, the first term of the divisor which gives 1

Algebraic long division Multiply x + 2 by 1 and subtract

Algebraic long division The remainder is –1. The quotient is 2x² - x + 1

You try one.

Do the next two

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). - 31 6 8 -2 1 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 - 1 3 1 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 41 0 - 4 0 6 1 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 12 48 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 192 198 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. - 24 8 -25 -50 4 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 - 25 50 0 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

You try one

Try another one

Do #’s 5 & 7

REMAINDER THEOREM When polynomial f(x) is divided by x – a, the remainder is f(a) f(x) = 2x 2 – 3x + 4 2 2 -3 4 2 4 1 2 6 Divide the polynomial by x – 2 Find f(2) f(2) = 2(2) 2 – 3(2) + 4 f(2) = 8 – 6 + 4 f(2) = 6

REMAINDER THEOREM f(x) = 3x 5 – 4x 3 + 5x - 3 Find f(-3) Try this one: Remember – Some terms are missing When synthetic division is used to evaluate a function, it is called SYNTHETIC SUBSTITUTION.

Do #’s 8 & 9

FACTOR THEOREM The binomial x – a is a factor of the polynomial f(x) if and only if f(a) = 0.

REMAINDER AND FACTOR THEOREMS Is x – 2 a factor of x 3 – 3x 2 – 4x + 12 2 1 -3 -4 12 1 2 -2 -6 -12 0 Yes, it is a factor, since f(2) = 0. Can you find the two remaining factors?

REMAINDER AND FACTOR THEOREMS (x + 3)( ? )( ? ) = x 3 – x 2 – 17x - 15 Find the two unknown ( ? ) quantities.

Find all the zeros. One factor has been given.

You try #’s 10 - 13

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