Sample
• How to calculate the x-intercepts of a graph (we equate the y-value to zero).
• Function notation, for example f(x).
◦ Always first take out a common factor if possible.
◦ Factorising can sometimes be done one or more times with the same expression.
◦ Always test your answer by multiplying out the factors again and ensuring that you obtain the original expression.
• Guidelines when you factorise:
◦ Determine how many terms must be factorised.
• If two terms (binomial) must be factorised:
• Firstly, take out a common factor, if possible, for example, 2 x 2 18 = 2( x 2 9).
• Check if further factorising is possible:
◦ Difference of two squares, for example, x 2 − 9:
◊ There are two factors (sets of brackets) with different signs, for example, (x − 3)(x + 3).
◊ Remember, the sum of two squares cannot be factorised.
◦ Difference of two cubes, for example, x 3 − 27:
◊ There are two factors (a binomial and a trinomial) with signs ( − )( + + ), for example, x 3 − 27 = (x − 3)( x 2 + 3x + 9).
◊ The first factor has the same sign as the original expression.
◊ The trinomial factor cannot be factorised again.
◦ Sum of two cubes, for example, x 3 + 27:
◊ There are two factors (a binomial and a trinomial) with signs ( + )(− + ), for example, x 3 + 27 = (x + 3)( x 2 − 3x + 9).
◊ The first factor has the same sign as the original expression.
◊ The trinomial factor cannot be factorised again.
• If three terms (trinomial) must be factorised:
◦ Firstly, take out a common factor if possible, for example,
2 x 2 − 8x + 6 = 2( x 2 − 4x + 3).
◦ Write the expression in the form a x 2 + bx + c if possible.
◦ Check whether further factorising is possible, for example, x 2 − 4x + 3 = (x − 1)(x − 3).
◦ If the sign of c is +: both factors (sets of brackets) will have the same sign:
• If b is +, the sign between the binomials of both factors will be +: (+)(+), for example, x 2 + 5x + 6 = (x + 2)(x + 3).
• If b is -, the sign between the binomials of both factors will be –: (–)(–), for example, x 2 4x + 3 = (x 1)(x 3).
◦ If the sign of c is : one factor (set of brackets) will be + and the other will be : (+)( ), for example, x 2 − 2x − 3 = (x + 1)(x − 3).
◦ Not all trinomials can be factorised.
◦ Test your answer by multiplying out again.
• If four terms (polynomial) must be factorised:
◦ Take out a common factor if possible, for example, 2xy + 2xb + 2ay + 2ab = 2(xy + xb + ay + ab).
◦ Group in groups of two terms each, for example, xy + xb + ay + ab = (xy + xb) + (ay + ab).
◦ Take out a common factor from each group if possible, for example, (xy + xb) + (ay + ab) = x(y + b) + a(y + b).
◦ Check whether common factors (sets of brackets) have formed, for example, x(y + b) + a(y + b).
◦ If common factors (sets of brackets) have not formed, regroup and try again.
◦ Once grouping has been successful, check whether further factorising is possible, for example, taking out a common factor:
x(y + b) + a(y + b) = (y + b)(x + a).
◦ Not all expressions with four terms can be factorised by grouping.
◦ Test your answer by multiplying out again.
Sample
◦ Groups can sometimes also consist of one and three terms or of three terms and one term.
• If five or more terms must be factorised:
◦ Take out a common factor, if possible.
◦ For five terms: group in groups of two and three terms, or of three and two terms.
◦ For six terms: group in groups of three and three terms, or of two and two and two terms.
◦ Group until common factors (sets of brackets) are obtained and factorise further if possible.
◦ Not all expressions of five and six terms can be factorised by grouping.
◦ Test your answer by multiplying out again.
Solving quadratic equations
• Write the equation in the standard form, in other words, a x 2 + bx + c = 0. Therefore, get all the terms on one side and equate them to zero, for example, write 2 x 2 8x = 6 as 2 x 2 8x + 6 = 0.
• Simplify the equation, if possible, for example:
◦ Divide by a common factor: 2 x 2 − 8x + 6 = 0 becomes x 2 − 4x + 3 = 0 by dividing by 2.
◦ Eliminate fractions by multiplying by the LCM of the denominators: 1 2 x 2 − x + 2 3 = 0 becomes 3 x 2 − 6x + 4 = 0 by multiplying by 6, the LCM of 2 and 3.
• Factorise fully by using some of the abovementioned types of factorising (mostly factorising of trinomials), for example, x 2 4x + 3 = 0 becomes (x − 1)(x − 3) = 0
• Apply the zero-product principle (if the product of factors is zero, at least one of the factors is also zero), for example, if (x 1)(x 3) = 0, then x 1 = 0 or x 3 = 0.
• Solve further, for example, if x − 1 = 0 or x − 3 = 0, then x = 1 or x = 3.
• Quadratic equations will usually have two answers (roots). Sometimes the two roots are identical, for example,
x 2 − 4x + 4 = 0
(x − 2)(x − 2) = 0
x = 2 OR x = 2
• If the trinomial cannot factorise into two factors (sets of brackets), the quadratic formula x = b ± √ b 2 − 4ac 2a must be used, for example, x 2 + 3x − 5 = 0
x = b ± √ b 2 − 4ac 2a = 3 ± √ (3) 2 − 4(1)( − 5) 2(1) = 3 ± √ 29 2 = 1,19 OR = − 4,19 (rounded to two decimal places)
• Function notation
◦ Names are sometimes given to functions, for example, f(x), g(x), k(x), and so on.
◦ f(x) = 2x + 1 means the function called f is defined by the equation y = 2x + 1.
◦ f(x) represents the y-values. Then f(3) is the corresponding y-value if the x-value is 3.
∴ f(3) = 2(3) + 1 = 7
Complete the revision exercise that follows.
Revision exercise
Sample
1. Fully factorise the following expressions: 1.1 2 x 16 − 2
2 x 2 − 11x + 12
14 x 2 y 4 − 21x y 3 + 7 x 4 y 2
3 x 3 − 18 x 2 − 2x + 12 1.5 8 x 3 − 4 x 2 − 27 y 3 − 6xy − 9 y 2 1.6 9 − x 2 + 10xy − 25 y 2 1.7 x 2(y − 3) + 7x(3 − y) − 10(3 − y) 1.8 (x + y) − (x − y) 2(x + y) 1.9 (x − y) 2 + (y − x) 1.10 x 3 + 2 x 2 − 2x − 4
2. Solve the following equations (to two decimal places where appropriate): 2.1 x 2 − 4x + 4 = 1 2.2 x 2 − 2(x + 5) = − 3x − 4 2.3 8 x 2 − 22x + 15 = 0 2.4 16 − (x − 2 ) 2 = 0 2.5 6 x 2 − x = 2 2.6 12 x 2 + 20x − 8 = 0
2.7 (x − 6)(x + 1) = 30 2.8 35 x 2 − 19x + 2 = 0
2.9 2 x 3 + 2 x 2 = 14x 2.10 x 2(3x − 2) − x(3x − 2) − 18x = − 12 2.11 2 x 3 − 4 x 2 + 9x = 0 2.12 x 3 − 5 x 2 − 12x = 0
2 x 16 − 2
( x 16 − 1 ) Common factor = 2( x 8 − 1)( x 8 + 1) Difference of squares = 2 (x 4 − 1 )(x 4 + 1)(x 8 + 1) Difference of squares
= 2 ( x 2 − 1)( x 2 + 1)( x 4 + 1)( x 8 + 1) Difference of squares
= 2(x − 1)(x + 1)(x 2 + 1)( x 4 + 1 )( x 8 + 1) Difference of squares 1.2 2 x 2 − 11x + 12
= (2x − 3 )(x − 4 )
Trinomial 1.3 14 x 2 y 4 − 21x y 3 + 7 x 4 y 2
= 7x y 2(2x y 2 − 3y + x 3)
Sample
1.4 3 x 3 − 18 x 2 − 2x + 12
= (3 x 3 − 18 x 2) + ( − 2x + 12)
= 3 x 2(x − 6) − 2(x − 6)
= (x − 6 )(3 x 2 − 2)
1.5 8 x 3 − 4 x 2 − 27 y 3 − 6xy − 9 y 2
= (8 x 3 − 27 y 3) + (−4 x 2 − 6xy − 9 y 2)
= (2x − 3y)(4 x 2 + 6xy + 9 y 2) − (4 x 2 + 6xy + 9 y 2)
Common factor
Grouping
Common factors
Common factor (x − 6 )
Grouping
Difference of cubes
= (4 x 2 + 6xy + 9 y 2)(2x − 3y − 1) Common factor (4 x 2 + 6xy + 9 y 2)
1.6 9 − x 2 + 10xy − 25 y 2
= 9 + ( x 2 + 10xy − 25 y 2)
Grouping for a trinomial
= 9 − (x 2 − 10xy + 25 y 2 ) Change sign for easier factorising
= 9 − (x − 5y)(x − 5y)
= 9 − (x − 5y) 2
= [3 − (x − 5y )][3 + (x − 5y )]
Factorise the trinomial
Difference of squares
= [3 − x + 5y ][3 + x − 5y] Remove inner sets of brackets
1.7 x 2(y − 3) + 7x (3 − y) − 10(3 − y)
= x 2(y − 3) − 7x(y − 3) + 10(y − 3 ) Change signs to get the common factor (y − 3)
= (y − 3)(x 2 − 7x + 10)
= (y − 3)( x − 2 )( x − 5 )
Common factor (y − 3)
Factorise the trinomial
1.8 (x + y ) − (x − y) 2(x + y)
= (x + y)[1 − (x − y) 2]
= (x + y)[1 − (x − y)][(1 + (x − y)]
1.9 (x − y) 2 + (y − x)
= (x − y) 2 − (x − y)
1.10 x 3 + 2 x 2 − 2x − 4
= (x 3 + 2 x 2) + ( 2x − 4)
= x 2(x + 2 ) − 2(x + 2)
= (x + 2 )(x 2 − 2)
2.1 x 2 − 4x + 4 = 1 x 2 − 4x + 3 = 0
(x − 3)(x − 1) = 0
Common factor (x + y)
Difference of squares
Remove inner sets of brackets
2.4 16 − (x − 2) 2 = 0
16 − (x 2 − 4x + 4) = 0
16 − x 2 + 4x − 4 = 0 − x 2 + 4x + 12 = 0
(x 2 − 4x − 12) = 0
Change sign of second term
x 2 − 4x − 12 = 0 (x − 6)(x + 2) = 0
Alternative method
Grouping
Common factors
Common factor (x + 2 )
Standard form of quadratic equation
Factorise the trinomial x = 3 OR x = 1
2.2 x 2 − 2(x + 5) = − 3x − 4
x 2 − 2x − 10 + 3x + 4 = 0 Remove brackets
x 2 + x − 6 = 0
(x + 3)(x − 2) = 0
2.3 8 x 2 − 22x + 15 = 0
(2x − 3)(4x − 5) = 0
Standard form of quadratic equation
Factorisethetrinomial x = − 3 OR x = 2
x = 3 2 OR x = 5 4
x = 6 OR x = 2
16 − (x − 2) 2 = 0
[4 − (x − 2)][4 + (x − 2)] = 0
(4 − x + 2)(4 + x − 2) = 0
Sample
Factorise the trinomial
Difference of squares
Remove inner sets of brackets
(6 − x)(2 + x) = 0 Simplify x = 6 OR x = 2
2.5 6 x 2 − x = 2
6 x 2 − x − 2 = 0
(3x − 2)(2x + 1) = 0
x = 2 3 OR x = 1 2
2.6 12 x 2 + 20x − 8 = 0
Standard form of quadratic equation
Factorise the trinomial
3 x 2 + 5x − 2 = 0 Divide by common factor 4
(3x − 1 )(x + 2) = 0
x = 1 3 OR x = − 2
Factorise the trinomial
Theme 7: Remainder and factor
2.7 (x − 6)(x + 1) = 30
We cannot apply the zero-product principle here, because the right-hand side ≠ 0
x 2 − 5x − 6 − 30 = 0 Remove brackets
x 2 − 5x − 36 = 0
(x − 9)(x + 4) = 0
x = 9 OR x = 4
2.8 35 x 2 − 19x + 2 = 0
(7x − 1)(5x − 2) = 0
x = 1 7 OR x = 2 5
2.9 2 x 3 + 2 x 2 − 14x = 0
2x(x 2 + x − 7) = 0
Standard form of quadratic equation
Factorise the trinomial
Factorise the trinomial
2x = 0 OR x = b ± √ b 2 − 4ac 2a Use quadratic formula for trinomial
2x = 0 OR x = − 1 + √ (1) 2 − 4(1)(− 7) 2(1) OR x = 1 − √ (1) 2 − 4(1)( − 7) 2(1)
x = 0 OR x = 2,19 OR x = − 3,19
2.10 x 2(3x − 2 ) − x(3x − 2) − 18x + 12 = 0 x 2(3x − 2 ) − x(3x − 2) − 6(3x − 2) = 0 Common factor 6 (3x − 2)(x 2 − x − 6 ) = 0 Common factor (3x − 2 ) (3x − 2)(x − 3)(x + 2) = 0 Factorise the trinomial x = 2 3 OR x = 3 OR x = − 2
2.11 2 x 3 − 4 x 2 + 9x = 0 x(2 x 2 − 4x + 9) = 0 Common factor
x = 0 OR 2 x 2 − 4x + 9 = 0
Sample
Trinomial cannot be factorised
x = 0 OR x = ( 5)± √ ( 5) 2 − 4 × 1 × (−12) 2 × 1
Trinomial cannot be factorised
Use quadratic formula for trinomial
x = 0 OR x = 5 ± √ 73 2 x = 0 OR x = 6,77 OR x = 1,77 2.13 2 3 x 2 + 8 3 x + 12 = 0 2 x 2 + 8x + 36 = 0 × LCM of 3 x 2 + 4x + 18 = 0 ÷ common factor of 2 x = 4 ± √ 16 − 4 × 1 × 18 2 Trinomial cannot be factorised
4 ± √ 16 − 72 2 = 4 ± √ 56 2 No real solution (roots are non-real) 2.14 4x + 1,5 = − x 2 4x + 3 2 = − x 2 Write 1,5 in fraction form 8x + 3 = − 2 x 2 × LCM of 2 2 x 2 + 8x + 3 = 0
Standard form of quadratic equation x = 8 ± √ 64 − 4 × 2 × 3 4 Use quadratic formula for trinomial
x = 0 OR x = ( 4) ± √ ( 4) 2 − 4 × 2 × 9 2(2) Use quadratic formula for trinomial x = 0 OR x = 4 ± √ 16 − 72 4 x = 0 OR x = 4 ± √ 54 4 x = 0 No further real solution The only real solution is x = 0. 2.12 x 3 − 5 x 2 − 12x = 0 x(x 2 − 5x − 12) = 0 Common factor x x = 0 OR x 2 5x − 12 = 0
x = 8 ± √ 40 4 x = − 0,42 OR x = − 3,58 2.15 2 5 x − 4 = 6 x 2 x 2 − 20x = 30 × LCM of 5x
2 x 2 − 20x − 30 = 0 Standard form of quadratic equation
x 2 − 10x − 15 = 0 ÷ common factor of 2 x = 10 ± √ 100 − 4 × 1 × ( − 15) 2 Use quadratic formula for trinomial x = 11,32 OR x = − 1,32 2.16 4 x(x − 2) = 2 x − 2 − 1 1 − x 4 x(x − 2)
4(x − 1) = 2x(x − 1) + x(x − 2) × LCM of
4x − 4 = 2 x 2 − 2x + x 2 − 2x Multiply brackets out 0 = 3 x 2 − 8x + 4 Standard form of quadratic equation 0 = (3x − 2)(x − 2) Factorise the trinomial x = 2 3 OR x = 2
Test roots: x ≠ 2,
Sample
(x − 4)(x + 2) = 0 Factorise x = 4 OR x =
1. REMAINDER THEO REM
In this subtheme, you will learn about the remainder theorem. As its name indicates, the remainder theorem deals with the remainder when a number or expression is divided by something.
If we divide a number like 21 by 5, we will get a remainder, because 5 is not a factor of 21. We can write it as: 21 5 = 4 remainder 1, or as 21 = 5 × 4 + 1, where:
• 5 is the divisor
• 4 is the quotient
• 1 is the remainder.
∴ 21 = divisor × quotient + remainder
The same applies when we divide an algebraic expression (polynomial) by another algebraic expression, for example, x 3 + 2 x 2 − 3x + 6 divided by x − 1. We can write it as x 3 + 2 x 2 − 3x + 6 = (x − 1)(x 2 + 3x) + 6, where:
• x 1 is the divisor
• x 2 + 3x is the quotient
• 6 is the remainder.
In general:
f(x) = divisor × quotient + remainder, where f(x) is an algebraic expression (polynomial) in terms of x.
We formulate the remainder theorem as follows:
If f(x) is divided by x − a until the remainder does not contain any x, then the remainder is f(a).
Applied to the example above: If we let f(x) = x 3 + 2 x 2 − 3x + 6, and f(x) is divided by x − 1 until the remainder does not contain any x, then the remainder is f(1).
Let us test this:
f(x) = x 3 + 2 x 2 − 3x + 6
∴ f(1) = (1) 3 + 2 (1) 2 − 3(1) + 6 = 1 + 2 − 3 + 6 = 6
Proof of the remainder theorem
*Not for exam purposes
Any polynomial f(x) can be written as f(x) = (x − a) × quotient + remainder
Then f(a) = (a − a) × quotient + remainder
= 0 × quotient + remainder
= 0 + remainder = remainder
Theme 7: Remainder and factor
Summary
• If f(x) is divided by x − a, then f(a) is the remainder.
How do we determine a? We let x − a = 0, and then solve for x.
Sample
For example, if f(x) = 2 x 3 + 5 x 2 − x + 3 is divided by x + 3, then we let x + 3 = 0 and solve for x. This gives x = − 3. Therefore, the remainder is f( 3).
• Likewise, if f(x) is divided by ax b, then we let ax b = 0, and solve for x. This gives x = b a . Then the remainder is f( b a ).
For example, if f(x) = 2 x 3 + 5 x 2 − x + 3 is divided by 2x + 3, we let 2x + 3 = 0 and solve for x. This gives x = − 3 2 . Therefore, the remainder is f( 3 2 ).
Worked example 1
If f(x) = x 2 + x + 2 is divided by x + 2, determine the remainder by using the remainder theorem.
Solution
Let x + 2 = 0
∴ x = − 2
f (−2) = ( 2) 2 + ( 2) + 2
= 4 − 2 + 2
= 4
∴ If f(x) is divided by x + 2, the remainder is 4.
Worked example 2
Determine the value of b if f(x) = x 3 − x 2 + 4x + b is divided by x − 1 and the remainder is 5.
Solution
Let x − 1 = 0
∴ x = 1
The remainder = f(1) = (1) 3 − (1) 2 + 4 (1) + b = 5
1 − 1 + 4 + b = 5
4 + b = 5
b = 1
∴ The value of b = 1.
Worked example 3
If f(x) = 4 x 3 + a x 2 + bx − 1 is divided by x − 1, the remainder is 6, and if f(x) is divided by x + 2, the remainder is − 27. Determine the values of a and b.
Solution
Remainder = f(1) = 4 (1) 3 + a (1) 2 + b(1) − 1 = 6
4 + a + b − 1 = 6
b = 3 − a ①
Let ① = ②: 3 − a = 2a − 3 6 = 3a a = 2
Sample
Remainder = f( − 2) = 4 (− 2 ) 3 + a (− 2 ) 2 + b(− 2) − 1 = − 27
32 + 4a − 2b − 1 = − 27
4a − 2b = 6
b = 2a − 3 ②
Substitute a = 2 back into ①: b = 3 − 2 b = 1
Exercise 1: Remainder theorem
1. Determine the remainder if f(x) = 4 x 2 + 14x + 18 is divided by x + 3 by using the remainder theorem.
2. If g(x) = 2 x 2 + 13x + 4 is divided by 2x + 3, determine the remainder by using the remainder theorem.
3. h(x) = 8 x 2 − 43x + 24 is divided by x − 5. Determine the remainder by using the remainder theorem.
4. What will the remainder be if f(x) = 2 x 3 − x 2 − 7x − 2 is divided by 2x − 3?
5. f(x) = 3 x 3 − 8 x 2 − 13x − 2. Determine the remainder if f(x) is divided by 3x + 5.
6. What will the remainder be if f(x) = 4 x 3 − 21 x 2 + 32x − 8 is divided by 4x − 5?
7. Determine the remainder if g(x) = 6 x 3 − 31 x 2 − 10x + 94 is divided by 3x− 5.
8. Which of x + 5, x − 2 or x + 4 will give a remainder of 25 if it is divided into f(x) = x 2 − 2x + 1?
9. Determine the remainder if g(x) = 9 x 3 − 9 x 2 − x + 1 is divided by 2x + 1.
10. Determine the remainder in terms of a if f(x) = a x 3 + 3 a 2 x 2 − 4ax + 5 is divided by 2x − a.
11. Determine the value of m if h(x) = 4 x 3 + 3 x 2 + mx + 2 is divided by 2x + 1 and the remainder is 4.
12. If f(x) = x 2 − 4x + 5 is divided by x − p, the remainder is 2. Determine the value(s) of p.
13. Determine the value of p if f(2) = 0 and it is given that f(x) = x 2 + 8p + 12.
14. If f(x) = a x 3 − b x 2 + 4x + 5 is divided by x − 1, the remainder is 11 and if f(x) is divided by 2x + 1, the remainder is 2. Determine the values of a and b.
15. The expression g(x) = x 3 + ax + b has a remainder of − 44 if it is divided by x + 3 and a remainder of 6 if it is divided by x − 2. Determine the values of a and b.
16. If g(x) = x 3 + p x 2 + 3 is divided by x + 2, the remainder is q. Determine p in terms of q.
17. What is the remainder if g(x) = 2 x 3 + 7 x 2 − m(x + 3) is divided by x − m?
18. If 3 x 3 − m x 2 − 6x + m and 5 x 3 − m x 2 − mx is divided by x − 1, their remainder is exactly the same. Determine the value of m.
2. FACTOR THEOREM
The factor theorem is very useful to factorise expressions (polynomials) of the third degree, amongst others.
The factor theorem is a special case of the remainder theorem, and states the following:
If f(x) is divided by x − a and the remainder is 0, then x − a is a factor of f(x).
In other words, if f(a) = 0, then x − a is a factor.
NOTE
The factor theorem only helps us to determine factors of the first degree (in other words, linear factors of the form x a or ax − b).
Factorising e x pressions of the third degree
Expressions of the third degree can be represented as a x 3 + b x 2 + cx + d.
Some expressions of the third degree can be factorised by using grouping, for example:
If grouping is not possible, we must use the factor theorem and determine a factor.
• If a factor x a needs to be found, f(a) must be equal to 0. Therefore, the remainder must be equal to zero.
• For example, to determine if x 1 is a factor of f(x), we determine whether f(1) = 0. If it does, then x 1 is a factor of f(x). If it does not, keep on trying to find a factor by using x + 1, then x 2, then x + 2, and so on, to test for a factor [i.e., determine whether f(−1) = 0, or f(2) = 0, or f( 2) = 0, and so on]. Therefore, keep on trying until the remainder is zero.
• For example, to determine whether 2x 1 is a factor of f(x), we determine whether f( 1 2 ) = 0. If it does, then 2x 1 is a factor of f(x).
• Once we find a factor, the factorisation process can start.
The following worked examples demonstrate factorising expressions of the third degree (polynomials) by using the factor theorem. There are four identifiable steps.
Worked example 4
Factorise f(x) = x 3 + 2 x 2 − 5x − 6 completely.
Solution
Step 1: Which possible factors can we investigate?
• This is determined by the factors of the coefficient of x 3 (in other words, the first term) and of the constant term (in other words, the last term).
• In this case, the coefficient of x 3 is just 1. In other words, the only factor is 1. The constant term is 6, in other words, the factors are 1, 2, 3 and 6.
• Possible factors are therefore (the easiest ones first):
◦ x − 1 we would therefore test whether f(1) = 0
◦ x + 1 we would therefore test whether f( 1) = 0
◦ x − 2 we would therefore test whether f(2) = 0
◦ x + 2 we would therefore test whether f( 2) = 0
◦ x − 3 we would therefore test whether f(3) = 0
◦ x + 3 we would therefore test whether f( 3) = 0
◦ x − 6 we would therefore test whether f(6) = 0
◦ x + 6 we would therefore test whether f( 6) = 0
Usually though, it is not necessary to go through the entire list to determine a factor!
Step 2: Test for a factor.
HINT
If we substitute 1 for x, in other words, determine f(1), the value of each term of f(1) is exactly the same as the coefficient of each term of f(x):
f(x) = x 3 + 2 x 2 − 5x − 6 and f(1) = 1 + 2 − 5 − 6
• Is x − 1 a factor?
Let x − 1 = 0
∴ x = 1
Sample
Determine f(1): f(1) = (1) 3 + 2 (1) 2 − 5(1) − 6 = 1 + 2 − 5 − 6 = − 8 ≠ 0
∴ x − 1 is not a factor, because the remainder is not zero.
• Is x + 1 a factor?
Let x + 1 = 0
x = − 1
Determine f( − 1): f( 1) = ( 1) 3 + 2 ( 1) 2 − 5( 1) − 6 = −
∴ x + 1 is a factor, because the remainder is zero.
Now we factorise further, to determine the other factors.
We already know that f(x) = (x + 1)(quotient).
Step 3: Determine the quotient (which is a trinomial, a x 2 + bx + c) if f(x) is divided by x + 1. There are several ways to do so. We use the so-called inspection method, which is shown below.
f(x) = (x + 1)(trinomial)
The first term of the trinomial: determine by what x in the first factor must be multiplied to get x 3:
x 3 + 2 x 2 − 5x − 6 = (x + 1)(x 2 + … + … ) x 3
∴ x × x 2 = x 3
The last term of the trinomial: determine by what + 1 must be multiplied to get 6: x 3 + 2 x 2 − 5x − 6 = (x + 1)( x 2 + … − 6)
∴ 1 × (−6) = − 6
The middle term of the trinomial: determine what is required to give us 2 x 2 (in x 3 + 2 x 2 − 5x − 6) once the brackets have been multiplied out and like terms simplified.
x 3 + 2 x 2 − 5x − 6 = (x + 1)( x 2 + … − 6) x 2
We already have x 2 when we multiply out, but we need 2 x 2, in other words, another x 2 .
The other x 2 needed, can be obtained by adding in an x in the middle term of the trinomial: x 2
x 3 + 2 x 2 − 5x − 6 = (x + 1) ( x 2 + x − 6) x 2
x 2 + x 2 = 2 x 2
We now test whether the third term of f(x) is correct:
The third term of f(x) is − 5x. When we multiply out: 6x
x 3 + 2 x 2 − 5x − 6 = ( x + 1)( x 2 + x − 6 ) x
6x + x = − 5x. Correct.
Step 4: Now factorise the trinomial, if possible.
x 3 + 2 x 2 − 5x − 6 = (x + 1)(x + 3)(x − 2) Worked example 5
Factorise completely the expression of the third degree, x 3 − 5x − 2.
Theme 7: Remainder and factor
Solution
Let f(x) = x 3 − 5x − 2.
Step 1: Which possible factors can we investigate?
Because the coefficient of x 3 is just 1, and the constant term is 2, only the following factors are possible:
Sample
• For x 3, there is only 1, and the factors of 2 is 1 and 2.
• Therefore, the possible factors are:
◦ x − 1 we would therefore test whether f(1) = 0
◦ x + 1 we would therefore test whether f( 1) = 0
◦ x − 2 we would therefore test whether f(2) = 0
◦ x + 2 we would therefore test whether f( 2) = 0
Step 2: Test for a factor.
• Is x 1 a factor?
Determine f(1): f(1) = 1 − 5 − 2 = − 6 ≠ 0
∴ x – 1 is not a factor, because the remainder is not zero.
• Is x + 1 a factor?
Let x + 1 = 0
∴ x = − 1
Determine f( − 1): f( − 1) = 2 ≠ 0
∴ x + 1 is not a factor, because the remainder is not zero.
• Is x + 2 a factor?
Let x + 2 = 0
∴ x = − 2
Determine f( − 2): f( − 2) = 0
∴ x + 2 is a factor of f(x)
∴ f(x) = (x + 2)(trinomial)
∴ x 3 − 5x − 2 = (x + 2)(a x 2 + bx + c)
Step 3: Determine the quotient a x 2 + bx + c, if x 3 − 5x − 2 is divided by x + 2.
The first term of the trinomial: determine by what must x be multiplied to get x 3 .
f(x) = (x + 2)( x 2 + … + … ) x 3
∴ x × x 2 = x 3
The last term of the trinomial: determine by what must 2 be multiplied to get 2.
f(x) = (x + 2)( x 2 + … 1) 2
∴ 2 × − 1 = − 2
The middle term of the trinomial:
f(x) = (x + 2)( x 2 + … 6)
2 x 2
We already have 2 x 2 for the second term of f(x), but there are no squares in the expression of f(x). Therefore, the 2 x 2 must now be cancelled by multiplying x by 2x in the trinomial.
2 x 2
f(x) = (x + 2)(x 2 − 2x − 1)
2 x 2
The 2 x 2 and 2 x 2 will cancel each other out.
∴ f(x) = (x + 2)( x 2 − 2x − 1)
Worked
example 6
Factorise 6 x 3 − 25 x 2 + 3x + 4 completely.
Sample
Multiply out to test whether the last two factors are correct.
Step 4: The trinomial cannot factorise further.
Solution
Let f(x) = 6 x 3 − 25 x 2 + 3x + 4
1. Which possible factors can we investigate?
Because the coefficient of x 3 is 6, and the constant term is 4, the following factors are possible:
• Factors of 6: 1, 2, 3 and 6, and the factors of 4: 1, 2 and 4.
Therefore, possible factors of f(x) = 6 x 3 − 25 x 2 + 3x + 4: ◦ x − 1 2
2. Test for factors:
Begin with the easiest ones (in other words, the first column above):
• Is x − 1 a factor?
Determine f(1): f(1) = − 12 ≠ 0
∴ x − 1 is not a factor, because the remainder is not zero.
• Is x + 1 a factor?
Let x + 1 = 0
∴ x = − 1
Determine f( − 1): f( − 1) = − 30 ≠ 0
∴ x + 1 is not a factor, because the remainder is not zero.
• Is x + 2 a factor?
Let x + 2 = 0
∴ x = − 2
Determine f( − 2): f( − 2) = − 150 ≠ 0
∴ x + 2 is not a factor, because the remainder is not zero.
• Is x 2 a factor?
Let x − 2 = 0
∴ x = 2
Determine f(2): f(2) = − 42 ≠ 0
∴ x − 2 is not a factor, because the remainder is not zero.
• Is x 3 a factor?
Let x − 3 = 0
∴ x = 3
Determine f(3): f(3) = − 50 ≠ 0
∴ x − 3 is not a factor, because the remainder is not zero.
• Is x + 3 a factor?
Let x + 3 = 0
∴ x = − 3
Determine f( − 3): f( − 3) = − 392 ≠ 0
∴ x + 3 is not a factor, because the remainder is not zero.
• Is x − 4 a factor?
Let x − 4 = 0
∴ x = 4
Determine f(4): f(4) = 0
Theme 7: Remainder and factor
3. Determine the quotient, a x 2 + bx + c, if 6 x 3 − 25 x 2 + 3x + 4 is divided by x − 4.
The first term of the trinomial: determine by what x must be multiplied to get 6 x 3 .
Sample
∴ x − 4 is a factor, because the remainder is zero.
∴ f(x) = (x − 4)(trinomial)
∴ 6 x 3 − 25 x 2 + 3x + 4 = (x − 4)(a x 2 + bx + c)
f(x) = (x − 4)(6 x 2 + … + … ) 6 x 3
∴ x × 6 x 2 = 6 x 3
The last term of the trinomial: determine by what must 4 be multiplied to get + 4.
f(x) = (x − 4)(6 x 2 + … − 1 ) 4
∴ − 4 × − 1 = 4
The middle term of the trinomial:
f(x) = (x − 4)(6 x 2 + … − 1) 24 x 2
We already have 24 x 2 for the second term of f(x), but we need 25 x 2 , therefore, we need another x 2 . − x 2
f(x) = (x − 4 )(6 x 2 − x − 1)
− 24 x 2
The sum of x 2 and 24 x 2 will give 25 x 2, which we need.
∴ f(x) = (x − 4)(6 x 2 − x − 1)
Multiply out to test whether the last two factors are correct.
4. The trinomial cannot factorise further.
• Revision exercises to refresh prior knowledge.
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• Index of mathematical terms.
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