Lesson 3- Polynomials
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Transcript of Lesson 3- Polynomials
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Objectives : - Definition - Dividing Polynomials
Next Lesson
- Factor Theorem - Remainder Theorem
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Polynomial
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kjxcxbxax nnn ..................21
Real numbers called coefficients Constant
n is the Degree of the polynomial
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Multiplying Polynomials
Expand all the terms
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126263)63)(2( 2232 xxxxxxxx
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Dividing Polynomials
This is trickier than multiplication
There are two main ways
─ Long Division─ By Inspection
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This PowerPoint presentation demonstrates two different methods of polynomial division.
Click here to see algebraic long division
Click here to see dividing “in your head”
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Divide 2x³ + 3x² - x + 1 by x + 2
3 22 2 3 1x x x x x + 2 is the divisor
The quotient will be here.
2x³ + 3x² - x + 1 is the dividend
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First divide the first term of the dividend, 2x³, by x (the first term of the divisor).
3 22 2 3 1x x x x
22xThis gives 2x². This will be the first term of the quotient.
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Now multiply 2x²by x + 2
3 22 2 3 1x x x x 3 22 4x x
22x
2xand subtract
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Bring down the next term, -x.
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
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Now divide –x², the first term of –x² - x, by x, the first term of the divisor
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
which gives –x.
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Multiply –x by x + 2
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
xand subtract
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Bring down the next term, 1
x
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
1
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Divide x, the first term of x + 1, by x, the first term of the divisor
13 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x 1which gives 1
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Multiply x + 2 by 1
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x
1
12x 1and subtract
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The remainder is –1.
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x
1
12x 1
The quotient is 2x² - x + 1
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Click here to see this example of algebraic long division again
Click here to see dividing “in your head”
Click here to end the presentation
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Divide 2x³ + 3x² - x + 1 by x + 2
When a cubic is divided by a linear expression, the quotient is a quadratic and the remainder, if any, is a constant.
Let the remainder be d.
Let the quotient by ax² + bx + c
2x³ + 3x² - x + 1 = (x + 2)(ax² + bx + c) + d
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2x³ + 3x² - x + 1 = (x + 2)(ax² + bx + c) + d
The first terms in each bracket give the term in x³
x multiplied by ax² gives ax³
so a must be 2.
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2x³ + 3x² - x + 1 = (x + 2)(2x² + bx + c) + d
The first terms in each bracket give the term in x³
x multiplied by ax² gives ax³
so a must be 2.
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2x³ + 3x² - x + 1 = (x + 2)(2x² + bx + c) + d
Now look for pairs of terms that multiply to give terms in x²
x multiplied by bx gives bx²
bx² + 4x² must be 3x²
2 multiplied by 2x² gives 4x²
so b must be -1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² + -1x + c) + d
Now look for pairs of terms that multiply to give terms in x²
x multiplied by bx gives bx²
bx² + 4x² must be 3x²
2 multiplied by 2x² gives 4x²
so b must be -1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² - x + c) + d
Now look for pairs of terms that multiply to give terms in x
x multiplied by c gives cx
cx - 2x must be -x
2 multiplied by -x gives -2x
so c must be 1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² - x + 1) + d
Now look for pairs of terms that multiply to give terms in x
x multiplied by c gives cx
cx - 2x must be -x
2 multiplied by -x gives -2x
so c must be 1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² - x + 1) + d
Now look at the constant term
2 multiplied by 1 gives 2
2 + d must be 1
then add d
so d must be -1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² - x + 1) - 1
Now look at the constant term
2 multiplied by 1 gives 2
2 + d must be 1
then add d
so d must be -1.
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2x³ + 3x² - x + 1 = (x + 2)(2x² - x + 1) - 1
The quotient is 2x² - x + 1 and the remainder is –1.
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Click here to see algebraic long division
Click here to see this example of dividing “in your head” again
Click here to end the presentation
21 April 2023 ML3 MH
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Do the following
1.
2.
3.
4.
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)12()7136( 23 xxxx
)3()15171392( 234 xxxxx
)47()21643283( 2234 xxxxxx
)1()23( 23 xxx
Exercises C1/C2 Page 82 Ex 3A, Nos 3, 6, 9, 16 to 20