CHAPTER 8.4
Special Case Factors!
Recognizing the binomial
product Difference of Squares
• Recall special case F.O.I.L. where middle
terms subtracts out.
• e.g. (a+b)(a-b)= a2 - b2
• When factoring Difference of Squares both
terms must be perfect squares, the sign must
be negative, and when you check by FOIL
the middle term subtracts out.
Factor A. x2 – 16 B. x2 – 10
C. z6 - 25
• A. (x)2- (4)2 Both terms are perfect
squares and sign is ( - ). = (x+4)(x-4)
• B. (x)2-(?)2 x is perfect square, but10 is
not, therefore NF over Integers.
• C. (z3)2-(5)2 Both terms are perfect
squares and sign is ( - ). = (z3+5)(z3-5)
Factor A. 25x2 – b2 B. 6x2 – 1
C. n8 - 36
• A. (5x)2- (b)2 Both terms are perfect
squares and sign is ( - ). = (5x+b)(5x-b)
• B. (?)2-(1)2 6x is not a perfect square,
but1 is, therefore NF over Integers.
• C. (n4)2-(6)2 Both terms are perfect
squares and sign is ( - ). = (n4+6)(n4-6)
Factor
z4 - 16
• (z2)2- (4)2 Is the difference of squares
and factors to (z2+4)(z2-4) always
continue to factor.
• (z2+4) (z2-4) second binomial factor is
also a difference of squares and
factors to(z-2)(z+2). (z2+4) Is simplified!
• (z2+4)(z+2)(z-2)
Factor
n4 - 81
• (n2)2- (9)2 Is the difference of squares
and factors to (n2+9)(n2-9) always
continue to factor.
• (n2+9) (n2-9) second binomial factor is
also a difference of squares and
factors to(n+3)(n-3).
• (n2+9) Is simplified!
• (n2+9)(n+3)(n-3)
Factoring Perfect Square Trinomials
• Recall patterns of special case FOIL
• (a+b)2 and
(a-b)2
• a2+2ab+b2 = (a+b)(a+b) = (a+b)2
• a2 - 2ab+b2 = (a -b)(a -b) = (a-b)2
Things to consider;
• Can the first and last terms be rewritten as
perfect squares?
• Does the middle term equal two times the
perfect squares of the first and last terms?
• If yes, then you can factor as a perfect
square trinomial.
Factor 9x2 – 30x +25
• 1st Rewrite first and last term as perfect
squares (3x)2 -30x + (5)2
• 2nd Does 2(a)(b)= middle term?
2(3x)(5)= 30x
• Then factor as a perfect square
trinomial.
•
(3x-5)(3x-5) = (3x-5) 2
Factor 16x2 +8x +1
• 1st Rewrite first and last term as perfect
squares (4x)2 +8x + (1)2
• 2nd Does 2(a)(b)= middle term?
2(4x)(1)= 8x
• Then factor as a perfect square
trinomial.
•
(4x+1)(4x+1) = (4x+1) 2
Factor 4x2 +37x +9
• 1st Rewrite first and last term as perfect
squares. (2x)2 +37x + (3)2
• 2nd Does 2(a)(b)= middle term?
2(2x)(3)= 12x = 37x
• Then factor is not a perfect square
trinomial. Factor by trial and error or
grouping.
Factor 4x2 +37x +9 by grouping.
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1st (a)(b) = (4)(9) = 36
2nd Consider factors of
3rd Rewrite middle terms
4th Factor by grouping
Check by F.O.I.L.
36
1, 36
2, 18
3, 12
4, 9
6, 6
4x2+36x + x +9
4x (x+9) +1(x+9)
(x+9)(4x+1)
4x2+37x+9
Factor x2 +14x +36
• 1st Rewrite first and last term as perfect
squares (x)2 +14x + (6)2
• 2nd Does 2(a)(b)= middle term?
2(x)(6)= 12x = 14x
• Then factor is not a perfect square
trinomial. Factor by trial and error or
grouping.
Factor x2 +14x +36 by grouping.
• 1st (a)(b) = (1)(36) = 36
• 2nd Consider factors of
36
•
1, 36
•
2, 18
•
3, 12
•
4, 9
•
6, 6
• None of the factors sum to 14, therefore
does not factor over integers.
Chapter 8.4 Objective 2
• While factoring, one must always ask
are the terms in simplest form? Could
the elements be factored further?
• This objective will answer that question,
and supply a check list for factoring
polynomials
Factoring Check List
• 1. Is there a common factor? If so, factor the
common factor out.
• 2. Is the polynomial the difference of two
squares? If so, factor.
• 3. Is the polynomial a perfect square
trinomial? If so, factor.
• 4. Is the polynomial a product of two
binomials? If so, factor.
• 5. Does the polynomial have four terms? If
so, try factoring by the grouping method.
Factor 3x2 -48
• 1st GCF 3(x2 -16)
• Continue to factor as difference of
squares.
• 3(x+4)(x-4)
• Factoring complete, check by FOIL.
Factor
x3 -3x2 -4x+12
• 1st There is not a GCF, but contains four
terms. Try factor by grouping.
x2(x3 -3x2 )-4(-4x+12)=
•
x2 (x-3)-4(x-3)=
•
(x-3)(x2 -4)= Continue to factor
•
(x-3)(x+2)(x-2)
• Factoring complete.
Factor 4x2y2 + 12xy2 +9y2
• 1st Factor GCF
• y2(4x2 +12x +9) Is remaining trinomial a
perfect square trinomial?
• First and last term can be written as
perfect squares. Two times first times
the last perfect squares[2(a)(b)] equals
middle term. Therefore, factors as.
•
y2(2x+3)2
NOW YOU TRY!
• 1. x4 -256
•
(x2 -16)(x+4)(x-4)
• 2. 9x2 -81
•
9(x+3)(x-3)
• 3. a2b – 3a2 -16b +48
•
(b-3)(a+4)(a-4)