Transcript x - msmccormick

MPM 2D
Factoring Polynomials
Example:
GCF = 4.
Factor 4x2 – 12x + 20.
= 4(x2 – 3x + 5)
Check the answer.
4(x2 – 3x + 5) = 4x2 – 12x + 20
A common binomial factor can be factored out of certain
expressions.
Example: Factor the expression 5(x + 1) – y(x + 1).
5(x + 1) – y(x + 1) = (x + 1) (5 – y)
Check. (x + 1) (5 – y) = 5(x + 1) – y(x + 1)
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Some polynomials can be factored by grouping terms to produce
a common binomial factor.
Notice the sign!
Examples: 1. Factor 2xy + 3y – 4x – 6.
2xy + 3y – 4x – 6 = (2xy + 3y) – (4x + 6) Group terms.
= y (2x + 3) – 2(2x + 3) Factor each pair of terms.
= (2x + 3) ( y – 2)
Factor out the common
binomial.
2. Factor 2a2 + 3bc – 2ab – 3ac.
2a2 + 3bc – 2ab – 3ac = 2a2 – 2ab + 3bc – 3ac
Rearrange terms.
= (2a2 – 2ab) + (3bc – 3ac) Group terms.
= 2a(a – b) + 3c(b – a)
Factor.
= 2a(a – b) – 3c(a – b)
= (a – b) (2a – 3c)
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b – a = – (a – b).
Factor.
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To factor a simple trinomial of the form x2 + bx + c, express the
trinomial as the product of two binomials. For example,
x2 + 10x + 24 = (x + 4)(x + 6).
Factoring these trinomials is based on reversing the FOIL process.
Example: Factor x2 + 3x + 2.
x2 + 3x + 2 = (x + a)(x + b)
Express the trinomial as a product of
two binomials with leading term x and
unknown constant terms a and b.
F O I
L
= x2 + bx + ax + ba
= x2 + (b + a)x + ba
= x2 + (1 + 2)x + 1 · 2
Apply FOIL to multiply the binomials.
Since ab = 2 and a + b = 3, it follows
that a = 1 and b = 2.
(Product-sum method)
Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
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Example: Factor x2 – 8x + 15.
x2 – 8x + 15 = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a + b = -8 and ab = 15.
It follows that both a and b are negative.
Negative Factors of 15
Sum
- 1, - 15
-15
-3, - 5
-8
x2 – 8x + 15 = (x – 3)(x – 5).
Check: (x – 3)(x – 5) = x2 – 5x – 3x + 15 = x2 – 8x + 15.
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13 + 36
Example: Factor x2 + 13x
36.
x2 + 13x + 36 = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a and b are:
two positive factors of 36
whose sum is 13.
Positive Factors of 36
Sum
1, 36
37
2, 18
3, 12
20
15
4, 9
6, 6
13
12
x2 + 13x + 36 = (x + 4)(x + 9)
Check: (x + 4)(x + 9) = x2 + 9x + 4x + 36 = x2 + 13x + 36.
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A polynomial is factored completely when it is written as a
product of factors that can not be factored further.
Example: Factor 4x3 – 40x2 + 100x.
4x3 – 40x2 + 100x
= 4x(x2 – 10x + 25)
= 4x(x – 5)(x – 5)
The GCF is 4x.
Use distributive property
to factor out the GCF.
Factor the trinomial.
Check: 4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25)
= 4x(x2 – 10x + 25)
= 4x3 – 40x2 + 100x
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Factoring complex trinomials of the form ax2 + bx + c, (a  1)
can be done by decomposition or cross-check method.
Example: Factor 3x2 + 8x + 4.
3  4 = 12
Decomposition Method
1. Find the product of
first and last terms
3. Rewrite the middle
term decomposed
into the two numbers
4. Factor by grouping
in pairs
2. We need to find factors of 12
whose sum is 8
1, 12
2, 6
3, 4
3x2 + 2x + 6x + 4
= (3x2 + 2x) + (6x + 4)
= x(3x + 2) + 2(3x + 2)
= (3x + 2) (x + 2)
3x2 + 8x + 4 = (3x + 2) (x + 2)
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Example: Factor 4x2 + 8x – 5.
4  5 = -20
We need to find factors of 20
whose difference is 8
Rewrite the middle term
decomposed into the
two numbers
Factor by grouping
in pairs
1, 20
-2, 10
4, 5
4x2 – 2x + 10x – 5
= (4x2 – 2x) + (10x – 5)
= 2x(2x – 1) + 5(2x – 1)
= (2x – 1) (2x + 5)
4x2 + 8x – 5 = (2x –1)(2x – 5)
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