# Polynomials

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```					Polynomials

A monomial in one variable is the product of a constant times a variable raised to a nonnegative integer power. Thus, a monomial is of the form

ax

k

where a is a constant, x is a variable, and k > 0 is an integer.

Monomial

Coefficient

Degree

3x

4

3

4 1

2x 9

2
-9

0

A polynomial in one variable is an algebraic expression of the form

an x n  an 1x n 1  a1x  a0

where an , an 1 ,, a1 , a0 are constants, called coefficients of the polynomial, n  0 is an integer, and x is a variable. If an  0, it is called the leading coefficient , and n is called the degree of the polynomial.

Determine the coefficients and degree of 2 x  3x  x  5.
4 2

Coefficients: 2, 0, -3, 1, -5 Degree: 4

2 x

3

    2 x  3x   x
2
3 3

 x  8 x  1  3x  5x  2
3
2




 8 x  5x   1  2

 5x  x  3x  1
3 2

2 x

3

 x  8 x  1  3x  5x  2
2 3

 

 2 x 3  x 2  8 x  1  3x 3  5x  2
 2 x  3x
3



3

 x

2

 8 x  5x   1  2

  x  x  13x  3
3 2

3x  2 x  4 x  3
2

 3x  x  3x  4 x   3x  3  2  x  2  4 x   2  3
2 2

 3x  12 x  9 x  2 x  8 x  6
3 2 2

 3x  14 x  17 x  6
3 2

Special Products
Difference of Two Squares

 x  a  x  a   x  a
2

2

Squares of Binomials, or Perfect Squares

 x  a   x  2ax  a 2 2 2  x  a   x  2ax  a
2 2

2

Special Products Miscellaneous Trinomials

 x  a  x  b  x  a  b x  ab 2 ax  bcx  d   acx  ad  bc x  bd
2

Cubes of Binomials, or Perfect Cubes

 x  a   x  3ax  3a x  a 3 3 2 2 3  x  a   x  3ax  3a x  a
3 3 2 2

3

Special Products Difference of Two Cubes

x  a   x  a  x  ax  a
3 3 2



2

 

Sum of Two Cubes

x  a   x  a  x  ax  a
3 3 2



2

Find the quotient and remainder when 2 x  8 x  x  4 is divided by 2 x  1
3 2 2

2 x  1 2 x  8x  x  4
2 3 2

x 2 3 2 2 x  1 2 x  8x  x  4

x 2 3 2 2 x  1 2 x  8x  x  4 2x
3

x

x 2 3 2 2 x  1 2 x  8x  x  4 2x
3 2

x  8x  2 x

x 4 2 3 2 2 x  1 2 x  8x  x  4 2x
3 2

x  8x  2 x  8x
2

4

x 4 2 3 2 2 x  1 2 x  8x  x  4 2x
3 2

x  8x  2 x  8x
2

4 2x  8

Check:

2 x

2

 1  x  4  2 x  8
3 2



 2 x  8x  x  4  2 x  8

 2 x  8x  x  4
3 2

Thus,

2 x  8x  x  4 2x  8  x4 2 2 2x  1 2x  1
3 2

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