4 3polynomial expressions

Polynomial Expressions
Frank Ma © 2011

A mathematics expression is a calculation procedure written in numbers, variables, and operation symbols.

Example A.
2 + 3x 

Example A.
2 + 3x  “the sum of 2 and 3 times x”

Example A.
4x2 – 5x

Example A.
4x2 – 5x  “the difference between 4 times the square of x and 5 times x”

Example A.
(3 – 2x)2

Example A.
(3 – 2x)2  “the square of the difference of 3 and twice x”

Example A.
An expression of the form #xN, where the exponent N is a
non-negative integer and # is a number, is called a monomial(one-term).

Example A.
An expression of the form #xN, where the exponent N is a non-negative integer and # is a number, is called a monomial(one-term).
For example, 3x2, –4x3, and 5x6 are monomials.

Example A.
Example B. Evaluatethe monomials if y = –4
a. 3y2

Example A.
a. 3y2
3y2 3(–4)2

Example A.
a. 3y2
3y2 3(–4)2
= 3(16) = 48

b. –3y2 (y = –4)

b. –3y2 (y = –4)
–3y2 –3(–4)2

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64)

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192
The sum of monomials are called polynomials (many-terms), these are expressions of the form
#xN ± #xN-1 ± … ± #x1 ± #
where # can be any number.

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192
#xN ± #xN-1 ± … ± #x1 ± #
For example, 4x + 7,

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192
#xN ± #xN-1 ± … ± #x1 ± #
For example, 4x + 7, –3x2 – 4x + 7,

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192
#xN ± #xN-1 ± … ± #x1 ± #
For example, 4x + 7, –3x2 – 4x + 7, –5x4 + 1are polynomials,

b. –3y2 (y = –4)
–3y2 –3(–4)2
= –3(16) = –48.
c. –3y3
–3y3 – 3(–4)3
= – 3(–64) = 192
#xN ± #xN-1 ± … ± #x1 ± #
For example, 4x + 7, –3x2 – 4x + 7, –5x4 + 1are polynomials,
1
is not a polynomial.
whereas the expression
x

Example C.Evaluate the polynomial 4x2 – 3x3 if x = –3.

The polynomial 4x2 – 3x3is the combination of two monomials; 4x2 and –3x3.

The polynomial 4x2 – 3x3is the combination of two monomials; 4x2 and –3x3. When evaluating the polynomial, we evaluate each monomial then combine the results.

Set x = (–3) in the expression,

Set x = (–3) in the expression, we get
4(–3)2 – 3(–3)3

4(–3)2 – 3(–3)3
= 4(9) – 3(–27)

4(–3)2 – 3(–3)3
= 4(9) – 3(–27)
= 36 + 81
= 117

4(–3)2 – 3(–3)3
= 4(9) – 3(–27)
= 36 + 81
= 117
Given a polynomial, each monomial is called a term.

4(–3)2 – 3(–3)3
= 4(9) – 3(–27)
= 36 + 81
= 117
#xN± #xN-1 ± … ± #x ± #
terms

4(–3)2 – 3(–3)3
= 4(9) – 3(–27)
= 36 + 81
= 117
#xN± #xN-1 ± … ± #x ± #
terms
Therefore the polynomial –3x2 – 4x + 7 has 3 terms,
–3x2, –4x and + 7.

Each term is addressed by the variable part.

Each term is addressed by the variable part. Hence the
x2-term of the –3x2 – 4x + 7 is –3x2,

x2-term of the –3x2 – 4x + 7 is –3x2, the x-term is –4x,

and the number term or the constant termis 7.

The number in front of a term is called the coefficient of that term.

The number in front of a term is called the coefficient of that term. So the coefficient of –3x2 is –3 .

Operations with Polynomials

Terms with the same variable part are called like-terms.

Like-terms may be combined.

For example, 4x + 5x = 9x

For example, 4x + 5x = 9x and 3x2 – 5x2 = –2x2.

Unlike terms may not be combined.

Unlike terms may not be combined. So x + x2 stays as x + x2.

Note that we write 1xN as xN , –1xN as –xN.

When multiplying a number with a term, we multiply it with the coefficient.

When multiplying a number with a term, we multiply it with the coefficient. Hence, 3(5x) = (3*5)x

When multiplying a number with a term, we multiply it with the coefficient. Hence, 3(5x) = (3*5)x =15x,

and –2(–4x) = (–2)(–4)x = 8x.

and –2(–4x) = (–2)(–4)x = 8x.
When multiplying a number with a polynomial, we may expand using the distributive law: A(B ± C) = AB ± AC.

Example D. Expand and simplify.
3(2x – 4) + 2(4 – 5x)

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
Polynomials in two or more variables.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
We form polynomials in two variables say, x & y, by adding
monomials of the form kx#y# where k is a number and the
powers are all nonnegative integers

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
powers are all nonnegative integers such as –5x3y2

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
powers are all nonnegative integers such as –5x3y2 or 3x2.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
Like–terms are terms where the variable parts are the same.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
For example 3x2y3 + 5x2y3 = 8x2y3

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
For example 3x2y3 + 5x2y3 = 8x2y3 but 3x2y3 + 5x3y3 can’t be
combined.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
combined. We evaluate them by assigning numbers to
x and/or y.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
x and/or y. If only one number is given, the result is a formula.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
If both numbers are given, then we get a numerical output.

3(2x – 4) + 2(4 – 5x)
= 6x – 12 + 8 – 10x
= –4x – 4
b. –3(x2 – 3x + 5) – 2(–x2 – 4x – 6)
= –3x2 + 9x – 15 + 2x2 + 8x +12
= –x2 + 17x – 3
If both numbers are given, then we get a numerical output.
We may do this for x, y and z or even more variables.

Example E. Expand and simplify.
3(2xy – 4x2y) + 2xy – 3xy2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
b. Evaluate 8xy – 8x2y – 3xy2 if x = 2.

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
c. Evaluate 8xy – 8x2y – 3xy2 if x = 2 and y = 3

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
We may put x = 2, y = 3 into the formula and do everything
all over

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
all over again or we may plug into y = 3 into part b which is
easier.

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
easier. We will do the easy way.

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
Input y = 3 into –16y – 6y2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
we get
–16(3) – 6(3)2

3(2xy – 4x2y) + 2xy – 3xy2
= 6xy – 8x2y + 2xy – 3xy2
= 8xy – 8x2y – 3xy2
Input x = 2, we get 8(2)y – 8(2)2y – 3(2)y2
= 16y – 32y – 6y2
= –16y – 6y2
we get
–16(3) – 6(3)2
= –48 – 54 = –102

4 3polynomial expressions

by math123a on Jun 22, 2011

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4 3polynomial expressions — Presentation Transcript