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Worksheet 2. 6 Factorizing Algebraic Expressions
Section 1 Finding Factors
Factorizing algebraic expressions is a way of turning a sum of terms into a product of smaller ones. The product is a multiplication of the factors. Sometimes it helps to look at a simpler case before venturing into the abstract. The number 48 may be written as a product in a number of different ways: 48 = 3 × 16 = 4 × 12 = 2 × 24 So too can polynomials, unless of course the polynomial has no factors (in the way that the number 23 has no factors). For example:
x^3 − 6 x^2 + 12x − 8 = (x − 2)^3 = (x − 2)(x − 2)(x − 2) = (x − 2)(x^2 − 4 x + 4)
where (x − 2)^3 is in fully factored form.
Occasionally we can start by taking common factors out of every term in the sum. For example,
3 xy + 9xy^2 + 6x^2 y = 3 xy(1) + 3xy(3y) + 3xy(2x) = 3 xy(1 + 3y + 2x)
Sometimes not all the terms in an expression have a common factor but you may still be able to do some factoring.
Example 1 : 9 a^2 b + 3a^2 + 5b + 5b^2 a = 3 a^2 (3b + 1) + 5b(1 + ba)
Example 2 : 10 x^2 + 5x + 2xy + y = 5 x(2x + 1) + y(2x + 1) Let T = 2x + 1 = 5 xT + yT = T (5x + y) = (2x + 1)(5x + y)
Example 3 : x^2 + 2xy + 5x^3 + 10x^2 y = x(x + 2y) + 5x^2 (x + 2y) = (x + 5x^2 )(x + 2y) = x(1 + 5x)(x + 2y)
Exercises:
- Factorize the following algebraic expressions: (a) 6x + 24 (b) 8x^2 − 4 x (c) 6xy + 10x^2 y (d) m^4 − 3 m^2 (e) 6x^2 + 8x + 12yx For the following expressions, factorize the first pair, then the second pair: (f) 8m^2 − 12 m + 10m − 15 (g) x^2 + 5x + 2x + 10 (h) m^2 − 4 m + 3m − 12 (i) 2t^2 − 4 t + t − 2 (j) 6y^2 − 15 y + 4y − 10
Section 2 Some standard factorizations
Recall the distributive laws of section 1.10.
Example 1 : (x + 3)(x − 3) = x(x − 3) + 3(x − 3) = x^2 − 3 x + 3x − 9 = x^2 − 9 = x^2 − 32
Example 2 : (x + 9)(x − 9) = x(x − 9) + 9(x − 9) = x^2 − 9 x + 9x − 81 = x^2 − 81 = x^2 − 92
Exercises:
- Expand the following, and collect like terms: (a) (x + 2)(x − 2) (b) (y + 5)(y − 5) (c) (y − 6)(y + 6) (d) (x + 7)(x − 7) (e) (2x + 1)(2x − 1) (f) (3m + 4)(3m − 4) (g) (3y + 5)(3y − 5) (h) (2t + 7)(2t − 7)
- Factorize the following: (a) x^2 − 16 (b) y^2 − 49 (c) x^2 − 25 (d) 4x^2 − 25
(e) 16 − y^2 (f) m^2 − 36 (g) 4m^2 − 49 (h) 9m^2 − 16
- Expand the following and collect like terms: (a) (x + 5)(x + 5) (b) (x + 9)(x + 9) (c) (y − 2)(y − 2) (d) (m − 3)(m − 3)
(e) (2m + 5)(2m + 5) (f) (t + 10)(t + 10) (g) (y + 8)^2 (h) (t + 6)^2
- Factorize the following: (a) y^2 − 6 y + 9 (b) x^2 − 10 x + 25 (c) x^2 + 8x + 16 (d) x^2 + 20x + 100
(e) m^2 + 16m + 64 (f) t^2 − 30 t + 225 (g) m^2 − 12 m + 36 (h) t^2 + 18t + 81
Section 3 Introduction to Quadratics
In the expression 5t^2 + 2t + 1, t is called the variable. Quadratics are algebraic expressions of one variable, and they have degree two. Having degree two means that the highest power of the variable that occurs is a squared term. The general form for a quadratic is
ax^2 + bx + c
Note that we assume that a is not zero because if it were zero, we would have bx + c which is not a quadratic: the highest power of x would not be two, but one. There are a few points to make about the quadratic ax^2 + bx + c:
- a is the coefficient of the squared term and a 6 = 0.
- b is the coefficient of x and can be any number.
- c is the called the constant term (even though a and b are also constant), and can be any number.
Quadratics may factor into two linear factors:
ax^2 + bx + c = a(x + k)(x + l)
where (x + k) and (x + l) are called the linear factors.
Exercises:
- Which of the following algebraic expressions is a quadratic? (a) x^2 − 3 x + 4 (b) 4x^2 + 6x − 1
(c) x^3 − 6 x + 2 (d) (^) x^12 + 2x + 1
(e) x^2 − 4 (f) 6x^2
Exercises:
- Factorize the following quadratics: (a) x^2 + 4x + 3 (b) x^2 + 15x + 44 (c) x^2 + 11x − 26 (d) x^2 + 7x − 30 (e) x^2 + 10x + 24
(f) x^2 − 14 x + 24 (g) x^2 − 7 x + 10 (h) x^2 − 5 x − 24 (i) x^2 + 2x − 15 (j) x^2 − 2 x − 15
The method that we have just described to factorize quadratics will work, if at all, only in the case that the coefficient of x^2 is 1. For other cases, we will need to factorize by
- Using the ‘ACE’ method, or by
- Using the quadratic formula
The ‘ACE’ method (pronounced a-c), unlike some other methods, is clear and easy to follow, as each step leads logically to the next. If you can expand an expression like (3x + 4)(2x − 3), then you will be able to follow this technique.
Example (^) Factorize 6x^2 − x − 12
1: Multiply the first term 6x^2 by the last term (−12) 2: Find factors of − 72 x^2 which add to −x. 3: Return to the original ex- pression and replace −x with − 9 x + 8x. 4: Factorize (6x^2 − 9 x) and (8x− 12). 5: One common factor is (2x − 3). The other factor, (3x + 4), is found by dividing each term by (2x − 3).
− 72 x^2 (− 9 x)(8x) = − 72 x^2 − 9 x + 8x = −x
6 x^2 − x − 12 = 6 x^2 − 9 x + 8x − 12 = 3 x(2x − 3) + 4(2x − 3) = (2x − 3)(3x + 4)
6: Verify the factorization by ex- pansion
(3x + 4)(2x − 3) = 3 x(2x − 3) + 4(2x − 3) = 6 x^2 − 9 x + 8x − 12 = 6 x^2 − x − 12
Example 3 : Factorize 4x^2 + 21x + 5.
- Multiply first and last terms: 4x^2 × 5 = 20x^2
- Find factors of 20x^2 which add to 21x and multiply to give 20x^2. These are 20 x and x.
- Replace 21x in the original expression with 20x + x: 4 x^2 + 21x + 5 = 4x^2 + 20x + x + 5
- Factorize the first two terms and the last two terms 4 x^2 + 20x + x + 5 = 4x(x + 5) + (x + 5)
the quadratic formula. Note that a = 1, b = 5, and c = 3.
x = −^5 ±^
so that the two roots are
k 1 = −5 +^
2 and^ k^2 =^
Then x^2 + 5x + 3 = (x − −5 +^
2 )(x^ −
Example 2 : Factorize 2x^2 − x − 5. Note that a = 2, b = −1, and c = −5. Then the solutions to 2x^2 − x − 5 = 0 are
x = −b^ ±
b^2 − 4 ac 2 a = 1 ±^
(−1)^2 − 4 × 2 × (−5)
2 × 2
So the two factors of 2x^2 − x − 5 are
(x − 1 +^
4 )^ and^ (x^ −^
and so the factorization is
2 x^2 − x − 5 = 2
x − 1 +^
x − 1 −
This right hand side of this equation should be expanded before it is believed!
Exercises:
- Factorize the following quadratics using the quadratic formula: (a) 3x^2 + 2x − 4 (b) x^2 + 3x + 1 (c) 2x^2 + 8x + 3 (d) 3x^2 + 5x + 1 (e) 3x^2 + 6x + 2
(f) 5x^2 + 7x − 2 (g) 3x^2 + 5x − 4 (h) 2x^2 + 4x + 1 (i) 5x^2 + 2x − 2 (j) 2x^2 + x − 7
Section 6 Uses of factorization
We can use factorization of expressions in a variety of ways. One way is to simplify algebraic fractions.
Example 1 : x^2 − 9 x − 3 =^
(x − 3)(x + 3) (x − 3) = x x^ −−^33 × (x + 3) = x + 3
Example 2 : x x^2 + 4x + 4 +^
x x + 2 =^
x (x + 2)^2 +^
x x + 2 = (^) (x + 2)x 2 + (^) x + 2x × x x^ + 2+ 2
= (^) (x + 2)x 2 + x
(^2) + 2x (x + 2)^2 = x
(^2) + x + 2x (x + 2)^2 = x (x(x + 2)^ + 3) 2
Another way of using factorization is in solving quadratic equations.
2. Simplify the following by first finding a common denominator:
Solve the following equations using the quadratic formula. Write the answers to two
- (h) x^22 x+6^2 −x^32 + - (i) x 33 x−−^927 x - (j) 2 xx 22 +−xx−−
- (b) x^4 −x 5 − x+2 (a) x+2^3 + x^5 +3x - (c) x x+1+2 + x x+3+
- (d) x 2 +5^6 x+6 + x 2 +8^2 x+ - (e) x 2 − 34 x− 10 − x 2 +5^1 x+ - (f) x 2 +6x+3x+9 − x+3
- (g) x x^22 +8+7xx+15+10 − x x+3+
- (h) x 2 x^2 −+6^9 − xx− - (j) x 23 +6xx − 2 xx+6+ (i) xx+1 − x^2 +3x
- (a) x^2 − 6 x + 8 = 3. Solve the following quadratic equations:
- (b) x^2 + 8x + 15 = - (c) x^2 + 7x + 12 =
- (d) x^2 + 9x − 22 = - (e) x^2 − 7 x + 12 = - (f) 2x^2 − x − 6 = - (g) 2x^2 − 13 x − 7 = - (h) 3x^2 − 10 x − 8 = - (i) 7x^2 + 13x − 2 = - (j) x^2 − 18 x + 77 =
- (a) x^2 − 3 x + 1 = decimal places.
- (b) 2x^2 − 6 x − 7 = - (c) 3x^2 + 2x − 2 = - (d) 2x^2 − 13 x + 7 =
Section 7 Multiplication and Division of Algebraic Fractions
We are often able to use factorization when we are multiplying or dividing algebraic expressions.
Example 1 : x^2 − 16 x + 3 ×^
x^2 + 5x + 6 x + 4 =^
(x + 4)(x − 4) x + 3 ×^
(x + 3)(x + 2) x + 4 = (x − 4)(x + 2)
Example 2 : 2 x^2 + 12x + 16 3 x^2 + 6x ×^
4 x^2 − 100 6 x + 30 =^
2(x^2 + 6x + 8) 3 x(x + 2) ×^
4(x^2 − 25) 6(x + 5) = 2(x 3 + 4)(x(x + 2)x^ + 2) × 4(x^ 6(+ 5)(x + 5)x^ −^ 5)
= 4(x^ + 4)( 9 xx^ −^ 5)
Example 3 : 6 x^2 + 9x x^2 + 8x + 15 ÷^
4 x + 6 x^2 − 9 =^
6 x^2 + 9x x^2 + 8x + 15 ×^
x^2 − 9 4 x + 6 = (^) (x^3 + 3)(x(2x^ + 3)x + 5) × (x^ 2(2+ 3)(x + 3)x^ −^ 3)
= (^3) 2(x(xx + 5)^ −^ 3)
Exercises:
- Simplify the following expressions: (a) 2 x x^22 −+2^5 xx− 3 ×^ x 22 x+4+1x (b) (^) x (^23) −x 7 +21x+12 × 49 xx−+63^12 (c) (^) xx (^2) +^2 +2x−x 20 × 2 x^25 −x+10^5 x−^12 (d) 3 x^25 −x−^1015 x− 8 ÷ 2 x x^22 −−^73 xx−^4 (e) x 4 x^2 − (^2) −^161 ÷ x 22 x+11 (^2) +5xx+28+
Answers 2.
Section 1
- (a) 6(x + 4) (b) 4x(2x − 1) (c) 2xy(3 + 5x) (d) m^2 (m^2 − 3)
(e) 2x(3x + 4 + 6y) (f) (4m + 5)(2m − 3) (g) (x + 5)(x + 2) (h) (m − 4)(m + 3)
(i) (t − 2)(2t + 1) (j) (2y − 5)(3y + 2)
Section 2
- (a) x^2 − 4 (b) y^2 − 25 (c) y^2 − 36
(d) x^2 − 49 (e) 4x^2 − 1 (f) 9m^2 − 16
(g) 9y^2 − 25
(h) 4t^2 − 49
- (a) (x + 4)(x − 4) (b) (y + 7)(y − 7) (c) (x + 5)(x − 5)
(d) (2x + 5)(2x − 5) (e) (4 + y)(4 − y) (f) (m + 6)(m − 6)
(g) (2m + 7)(2m − 7)
(h) (3m + 4)(3m − 4)
- (a) x^2 + 10x + 25 (b) x^2 + 18x + 81 (c) y^2 − 4 y + 4
(d) m^2 − 6 m + 9 (e) 4m^2 + 20m + 25 (f) t^2 + 20t + 100
(g) y^2 + 16y + 64
(h) t^2 + 12t + 36
- (a) (y − 3)^2 (b) (x − 5)^2 (c) (x + 4)^2
(d) (x + 10)^2 (e) (m + 8)^2 (f) (t − 15)^2
(g) (m − 6)^2
(h) (t + 9)^2
Section 3
- (a), (b), (e), and (f)
Section 4 part 1
- (a) (x + 3)(x + 1) (b) (x + 11)(x + 4) (c) (x + 13)(x − 2) (d) (x + 10)(x − 3)
(e) (x + 6)(x + 4) (f) (x − 12)(x − 2) (g) (x − 5)(x − 2)
(h) (x − 8)(x + 3) (i) (x + 5)(x − 3)
(j) (x − 5)(x + 3)
Section 4 part 2
- (a) (2x + 3)(x + 4) (b) (3x + 1)(x + 5) (c) (2x + 3)(3x + 4) (d) (2x + 5)(x + 2) (e) (3x + 2)(4x + 1)
(f) (2x + 1)(x − 3) (g) (3x + 2)(x − 4) (h) (3x + 4)(x − 5) (i) (5x + 2)(x + 3) (j) (5x + 2)(2x + 3)
Section 5
- (a) 3(x − −2+
√ 52 6 )(x^ −^ −^2 −
√ 52 6 ) (b) (x − −3+
√ 5 2 )(x^ −^ −^3 −
√ 5 2 ) (c) 2(x − −8+
√ 40 4 )(x^ −^ −^8 −
√ 40 4 ) (d) 3(x − −5+
√ 13 6 )(x^ −^ −^5 −
√ 13 6 ) (e) 3(x − −6+
√ 12 6 )(x^ −^ −^6 −
√ 12 6 )
(f) 5(x − −7+
√ 89 10 )(x^ −^ −^7 −
√ 89 10 ) (g) 3(x − −5+
√ 73 6 )(x^ −^ −^5 −
√ 73 6 ) (h) 2(x − −4+
√ 8 4 )(x^ −^ −^4 −
√ 8 4 ) (i) 5(x − −2+
√ 44 10 )(x^ −^ −^2 −
√ 44 10 ) (j) 2(x − −1+
√ 57 4 )(x^ −^ −^1 −
√ 57 4 )
Section 6
- (a) x (b) 3 x^2 x−^4
(c) x+1 3 (d) x x+2+
(e) x− 24 (f) x− 63
(g) x x+5+ (h) 2( xx+2−4)
(i) x 32 (j) (^2) xx+3+
- (a) (^5) (xx+2)(^2 +13xx+3)+ (b) 2(2 (x+2)(x^2 +3xx−+5)5) (c) 2( (xx+2)(^2 +5xx+4)+5)
(d) (^) (x+2)(2(4xx+3)(+17)x+5) (e) (^) (x+2)(^3 xx+17+3)(x−5) (f) (^) x−+3^1 (g) 0
(h) −x 2(^2 −x−^6 x3)+ (i) (^) (x−+1)(x(x−x+3)1) (j) − (2(x+6)x−1)
- (a) 4, 2 (b) −5, − 3
(c) −4, − 3 (d) −11, 2
(e) 4, 3 (f) −2, (^32)
(g) 7, − (^12) (h) 4, − (^23)
(i) −2, (^17) (j) 11, 7
- (a) 3 ±
√ 5 2 (b)^32 ±^
√ 92 4 (c)^ −^13 ±^
√ 28 6 (d)^134 ±^
√ 113 4
