MA 1128: Lecture 10 – 10/17/16
Addition, Subtraction, and Multiplication
of Polynomials
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Polynomials
A polynomial in one variable x is an expression which is the sum of only
constant terms, x-terms, x2-terms, x3-terms, etc.
For example, 7x4 – 2x2 + x + 3 is a polynomial.
The variable doesn’t have to be x, of course, and the book talks about
polynomials in several variables.
In several variables, we might have y3-terms and x2y4-terms.
We’ll stick to one variable, for the most part.
You could say that a “nomial” is the same as a term, so “polynomial” means many
terms.
We’ll actually take “poly” to mean any number of terms, so 3x2 by itself would
be a polynomial.
Occasionally, the words monomial, binomial, and trinomial are used to explicitly
describe a polynomial with one, two, or three terms.
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Adding Polynomials
We’ll need to be able to add and subtract polynomials.
We’ve done this already, and called it simplifying.
Pay special attention to combining like terms.
Example. (3x2 – 2x + 1) + (7x3 – 2x2 + x + 1).
With addition, the parentheses don’t really do anything, since we can add the
terms in any order. Dropping the parentheses, we get
= 3x2 – 2x + 1 + 7x3 – 2x2 + x + 1
Combining like terms we get:
= 7x3 + x2 – x + 2
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Practice Problems
1.
Add the polynomials 4x4 – 2x2 + x + 7 and 3x3 – 2x + 8.
Click for answers:
4x4 + 3x3  2x2 – x + 15
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Subtraction of Polynomials
Example. (7x2 + 2x – 1) – (3x2 – 2x + 8).
Here the second set of parentheses are doing something very important.
They tell us that every term inside is being subtracted.
We must subtract 3x2, subtract 2x, and subtract 8.
= 7x2 + 2x – 1 – 3x2 + 2x – 8
Note that the signs changed on each term in the second polynomial.
= 4x2 + 4x – 9.
Example. (5x2 – 2x + 7) – (10x3 – 7x2 + 8x – 13)
= 5x2 – 2x + 7 – 10x3 + 7x2 – 8x + 13
= 10x3 + 12x2 – 10x + 20.
Next Slide
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Practice Problems
1.
2.
Subtract: (x2 + 2x – 3) – (7x – 4).
Subtract: (x2 – 4) – (3x3 – 2x2 + x – 7).
Answers:
1) x2 – 5x + 1
2) 3x3 + 3x2 – x + 3
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Multiplying Polynomials
We’ve done a little of this, but not too much.
Consider the multiplication (x2 – 2x + 2)(x + 1).
Recall that multiplication distributes over addition.
A careful application of this concept leads to the fact that
Every term in the first polynomial needs to be multiplied times every term in
the second.
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Example
In (x2 – 2x + 2)(x + 1),
each of the terms x2, 2x, and 2
must be multiplied times each of the terms x and 1.
Let’s be systematic: x2 times x and x2 times 1
2x times x and 2x times 1
2 times x and 2 times 1.
This gives us the new terms: x3 and x2
2x2 and 2x
2x and 2
These get added together: x3 + x2 – 2x2 – 2x + 2x + 2
= x3 – x2 + 0 + 2
= x3 – x2 + 2
Next Slide
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Example
3x2 times everything
x times everything
(3x2 – x)(x2 – 2x + 3)
= (3x2)(x2) + (3x2)(-2x) + (3x2)(3)
+ (-x)(x2) + (-x)(-2x) + (-x)(3)
= 3x4 – 6x3 + 9x2
– x3 + 2x2 – 3x
= 3x4 – 7x3 + 11x2 – 3x.
Next Slide
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Example
(x2 + 2x – 1)(2x2 – 3x + 2)
= (x2)(2x2) + (x2)(-3x) + (x2)(2)
+ (2x)(2x2) + (2x)(-3x) + (2x)(2)
+ (-1)(2x2) + (-1)(-3x) + (-1)(2)
= 2x4 – 3x3 + 2x2
+ 4x3 – 6x2 + 4x
– 2x2 + 3x – 2
= 2x4 + x3 – 6x2 + 7x – 2
Next Slide
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Practice Problems
1.
2.
3.
(x2 + 3)(x – 2)
(x2 + 2x – 1)(x + 7)
(x2 – x + 2)(x2 + 2x – 3)
Answers:
1) x3 – 2x2 + 3x – 6
2) x3 + 9x2 + 13x – 7
3) x4 + x3 – 3x2 + 7x – 6
End
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