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Dividing polynomials
This PowerPoint presentation
demonstrates two different methods
of polynomial division.
© Kristen A. Treglia
2007
Algebraic long division
Divide 2x³ + 3x² - x + 1 by x + 2
x  2 2x 3
 3x 2
 x  1
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).
2x 2
x  2 2x 3
 3x 2
 x  1
This gives 2x².
This will be
the first term
of the
quotient.
Algebraic long division
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
2x 2
x 2
Now multiply
2x² by x +
2
and
subtract
Algebraic long division
Bring down the
next term, -x.
2x 3
 4x 2
2x 2
x  2 2x 3
 3x 2
 x  1
x 2
x
Algebraic long division
2x 3
 4x 2
x 2
x
2x 2
x
x  2 2x 3
 3x 2
 x  1
Now divide –x²,
the first term of
–x² - x, by x, the
first term of
the divisor
which gives –x.
Algebraic long division
x 2
x
2x 2
x
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
x 2
 2x
x
Multiply –x by x + 2
and subtract
Algebraic long division
Bring down the
next term, 1
x 2
x
2x 2
x
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
x 2
 2x
x 1
Algebraic long division
Divide x, the first
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
x 2
x
2x 2
x 1
x 2
 2x
x 1
term of x + 1, by x,
the first term of
the divisor
which gives 1
Algebraic long division
x 2
x
x 2
 2x
2x 2
x
1
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
x 1
x  2
1
Multiply x + 2 by 1
and subtract
Algebraic long division
The remainder is –1.
x  2 2x 3
 3x 2
 x  1 2x
3
 4x 2
x 2
x
x 2
 2x
2x 2
x 1
x 1
x  2
1
The quotient is
2x² - x + 1

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dividing-polynomials-slide-share-1196832292862534-5-converted.pptx

  • 1. Dividing polynomials This PowerPoint presentation demonstrates two different methods of polynomial division. © Kristen A. Treglia 2007
  • 2. Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x  2 2x 3  3x 2  x  1 x + 2 is the divisor The quotient will be here. 2x³ + 3x² - x + 1 is the dividend
  • 3. Algebraic long division First divide the first term of the dividend, 2x³, by x (the first term of the divisor). 2x 2 x  2 2x 3  3x 2  x  1 This gives 2x². This will be the first term of the quotient.
  • 4. Algebraic long division x  2 2x 3  3x 2  x  1 2x 3  4x 2 2x 2 x 2 Now multiply 2x² by x + 2 and subtract
  • 5. Algebraic long division Bring down the next term, -x. 2x 3  4x 2 2x 2 x  2 2x 3  3x 2  x  1 x 2 x
  • 6. Algebraic long division 2x 3  4x 2 x 2 x 2x 2 x x  2 2x 3  3x 2  x  1 Now divide –x², the first term of –x² - x, by x, the first term of the divisor which gives –x.
  • 7. Algebraic long division x 2 x 2x 2 x x  2 2x 3  3x 2  x  1 2x 3  4x 2 x 2  2x x Multiply –x by x + 2 and subtract
  • 8. Algebraic long division Bring down the next term, 1 x 2 x 2x 2 x x  2 2x 3  3x 2  x  1 2x 3  4x 2 x 2  2x x 1
  • 9. Algebraic long division Divide x, the first x  2 2x 3  3x 2  x  1 2x 3  4x 2 x 2 x 2x 2 x 1 x 2  2x x 1 term of x + 1, by x, the first term of the divisor which gives 1
  • 10. Algebraic long division x 2 x x 2  2x 2x 2 x 1 x  2 2x 3  3x 2  x  1 2x 3  4x 2 x 1 x  2 1 Multiply x + 2 by 1 and subtract
  • 11. Algebraic long division The remainder is –1. x  2 2x 3  3x 2  x  1 2x 3  4x 2 x 2 x x 2  2x 2x 2 x 1 x 1 x  2 1 The quotient is 2x² - x + 1