THEME 7
REMAINDER AND FACTOR THEOREMS
CAPS learning requirements
Learners should know and be able to apply the following:
1. Factorise polynomials of the third degree.
2. Apply the remainder and factor theorems to polynomials of not more than the third degree (no proofs are required).
Term 2
Duration 1 week
Paper 1
Weight
Facilitator tips
This theme forms part of differential calculus, of which the weight is 35 ± 3 marks of Paper 1.
• When learners factorise, they must first use the general rules of factorising, namely:
◦ Always first take out a common factor if possible.
◦ Factorising can sometimes be done one or more times with the same expression until the expression is fully factorised.
◦ The number of terms usually determines which type of factorising to use.
• When the roots of a function must be calculated, the function must be equated to zero.
• When a quadratic function cannot be factorised, the quadratic formula must be used to calculate the roots.
Sample
◦ When the factor theorem is used, learners must always begin by testing for the easiest factors that are possible, for example, x − 1; x + 1; and so on.
• Answers must always be tested by multiplying out the factors and ensuring that the original expression is obtained.
• The sum of two squares cannot be factorised, but the sum of two cubes can.
• The order of factors (brackets) does not matter.
• When a polynomial of the third degree is factorised, any of the three factors (if there are three) can be used as the first factor.
Introduction
In this theme, learners will learn more about:
• the remainder that arises when an algebraic expression (polynomial) is divided by a linear expression.
• factorising expressions of the third degree.
• solving equations of the third degree.
Prior knowledge
To master this theme, learners should already know the following:
• What the concepts ‘factor’, ‘divisor’, ‘quotient’ and ‘remainder’ mean.
• How to factorise by using different methods:
◦ Common factor
◦ Difference of squares
◦ Grouping
◦ Trinomial
◦ Sum and difference of cubes
• How to solve quadratic equations by getting all the terms on one side of the equals sign (=), and zero on the other side (solutions of an equation are also called roots).
• How to calculate the x-intercepts of a graph (we equate the y-value to zero).
• Function notation, for example f(x).
We will first revise factorising and the solving of quadratic equations as we dealt with in Grades 10 and 11.
Factorising of polynomials
• Three general rules to remember:
◦ 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 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).
Sample
◊ 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).
◦ 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.
◦ Groups can sometimes also consist of one and three terms or of three terms and one term.
◦ 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.
• 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
• 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
Sample
• 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
Learners complete the revision exercise below.
• 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.
1. Fully factorise the following e
pressions:
2. Solve the following equations (to two decimal places where appropriate):
3. The function g(x) = 2 x 2 4x + 3 is given. Determine: 3.1 g( 1) 3.2 g( 1 2 ) 3.3 x if g(x) = 19 4. The function g(x) = x 3 + 2 x 2 4x + 3 is given. Determine: 4.1 g(2) 4.2 g( 3) Solutions 1.1 2 x 16 − 2
= 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 )
1.3 14 x 2 y 4 − 21x y 3 + 7 x 4 y 2
= 7x y 2(2x y 2 − 3y + x 3)
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)
Trinomial
1.8 (x + y ) − (x − y) 2(x + y) = (x + y)[1 − (x − y) 2]
= (x + y)[1 − (x − y)][(1 + (x − y)]
Common factor
Grouping
Common factors
Common factor (x − 6 )
Grouping
= (2x − 3y)(4 x 2 + 6xy + 9 y 2) − (4 x 2 + 6xy + 9 y 2) 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
Factorise the trinomial
= [3 − (x − 5y )][3 + (x − 5y )] Difference of squares
= [3 − x + 5y ][3 + x − 5y]
1.7 x 2(y − 3) + 7x (3 − y) − 10(3 − y)
Sample
Remove inner sets of brackets
= 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
Common factor (x + y)
Difference of squares = (x + y)[1 − x + y][1 + x − y]
Remove inner sets of brackets 1.9 (x − y) 2 + (y − x) = (x − y) 2 − (x − y)
Change sign of second term
Common factor (x − y) 1.10 x 3 + 2 x 2 − 2x − 4 = (x 3 + 2 x 2) + ( 2x − 4)
Grouping = x 2(x + 2 ) − 2(x + 2)
Common factors = (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 + 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
Factorise the trinomial x = − 3 OR x = 2
Factorise the trinomial x = 3 2 OR x = 5 4
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
x 2 − 4x − 12 = 0
(x − 6)(x + 2) = 0 x = 6 OR x = − 2
Alternative method
16 − (x − 2) 2 = 0
[4 − (x − 2)][4 + (x − 2)] = 0 Difference of squares
(4 − x + 2)(4 + x − 2) = 0
Remove inner sets of brackets
(6 − x)(2 + x) = 0 Simplify x = 6 OR x = − 2
2.7 (x − 6)(x + 1) = 30
2.5
x − 2 = 0 Standard form of quadratic equation
(3x − 2)(2x + 1) = 0 Factorise the trinomial x = 2 3 OR x = 1 2
2.6 12 x 2 + 20x − 8 = 0
3 x 2 + 5x − 2 = 0 Divide by common factor 4
(3x − 1 )(x + 2) = 0
Factorise the trinomial x = 1 3 OR x = − 2
Theme 7: Remainder and factor
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
Sample
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
2x = 0 OR x = b ± √ b 2 − 4ac 2a
Standard form of quadratic equation
Factorise the trinomial
Factorise the trinomial
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
x = 0 OR 2 x 2 − 4x + 9 = 0
Common factor
Trinomial cannot be factorised
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
x = 0 OR x 2 5x − 12 = 0
x = 0 OR x = ( 5)± √ ( 5) 2 − 4 × 1 × (−12) 2 × 1
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
Solution: x = 2 3 Sample
Common factor x
Trinomial cannot be factorised
Use quadratic formula for trinomial
of
formula
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
(x − 2) =
x − 2 + 1 x − 1 Take out – from last term and change the sign
4(x − 1) = 2x(x − 1) + x(x − 2) × LCM of x(x − 2)(x − 1) 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, as this will cause division by zero
3.1 g(x) = 2 x 2 − 4x + 3
g(− 1) = 2 (− 1) 2 − 4(− 1) + 3
3.2 g(x) = 2 x 2 − 4x + 3
Substitute -1 for x
3.3 2 x 2 − 4x + 3 = 19 Let g(x) = 19
2 x 2 − 4x 16 = 0 Standard form
4.1 g(x) = x 3 + 2 x 2 − 4x + 3 g(2) = 8 + 8 − 8 + 3
THEO REM
In this subtheme, learners 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.
Theme 7: Remainder and factor
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:
Sample
• 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
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.
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
∴
f (−2) = (− 2) 2 + (− 2) + 2 = 4 − 2 + 2 = 4
Sample
∴ 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
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
Let ① = ②: 3 − a = 2a − 3
6 = 3a a = 2
Substitute a = 2 back into ①: b = 3 − 2
b = 1
E x ercise 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.
Let x + 3 = 0
∴ x = − 3
Remainder = f(−3) = 4 ( 3) 2 + 14(−3) + 18
Substitute 3 for x
= 4(9) + 14(−3) + 18 Simplify = 12
2. If g(x) = 2 x 2 + 13x + 4 is divided by 2x + 3, determine the remainder by using the remainder theorem.
Let 2x + 3 = 0
∴ x = − 3 2
Theme 7: Remainder and factor theorems
3. h(x) = 8 x 2 − 43x + 24 is divided by x − 5. Determine the remainder by using the remainder theorem.
Let x − 5 = 0 ∴ x = 5
Sample
Remainder = g( 3 2 ) = 2 ( 3 2 ) 2 + 13( 3 2 ) + 4
Substitute 3 2 for x = − 11
Remainder = h(5) = 8 (5) 2 − 43(5) + 24 = 9 Substitute 5 for x
4. What will the remainder be if f(x) = 2 x 3 − x 2 − 7x − 2 is divided by 2x − 3?
Let 2x − 3 = 0 ∴ x = 3 2
Remainder = f( 3 2 ) = 2 ( 3 2 ) 3 − ( 3 2 ) 2 − 7( 3 2 ) − 2 = – 8 Substitute 3 2 for x
5. f(x) = 3 x 3 − 8 x 2 − 13x − 2.
Determine the remainder if f(x) is divided by 3x + 5.
Let 3x + 5 = 0
∴ x = 5 3
Remainder = f( 5 3 ) = 3 ( 5 3 ) 3 − 8 ( 5 3 ) 2 − 13( 5 3 ) − 2 Substitute 5 3 for x = − 148 9
6. What will the remainder be if f(x) = 4 x 3 − 21 x 2 + 32x − 8 is divided by 4x − 5?
Let 4x − 5 = 0
∴ x = 5 4
Remainder = f( 5 4 ) = 4 ( 5 4 ) 3 − 21 ( 5 4 ) 2 + 32( 5 4 ) − 8 = 7 Substitute 5 4 for x
7. Determine the remainder if g(x) = 6 x 3 − 31 x 2 − 10x + 94 is divided by 3x− 5.
Let 3x − 5 = 0
∴ x = 5 3
Remainder = g( 5 3 ) = 6 ( 5 3 ) 3 − 31 ( 5 3 ) 2 − 10( 5 3 ) + 94 = 19
Substitute 5 3 for x
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?
Let x + 5 = 0
∴ x = − 5
f( 5) = 36
Let x − 2 = 0 ∴ x = 2
f(2) = 1
Let x + 4 = 0
∴ x = − 4
f( 4) = 25
∴ x + 4 will give a remainder of 25.
Substitute 5 for x
10. Determine the remainder in terms of a if f(
Substitute 2 for x
Substitute 4 for x
9. Determine the remainder if g(x) = 9 x 3 − 9 x 2 − x + 1 is divided by 2x + 1.
Let 2x + 1 = 0
a
is divided by 2x − a. Let 2x − a = 0
∴ x = a 2
Sample
Remainder = f( a 2 ) = a ( a 2 ) 3 + 3 a 2 ( a 2 ) 2 − 4a( a 2 ) + 5 Substitute a 2 for x = a( a 3 8 ) + 3 a 2( a 2 4 ) − 4( a 2 2 ) + 5 Simplify = ( a 4 8 ) + 3( a 4 4 ) − 2( a 2) + 5
= ( a 4 8 ) + 6( a 4 8 ) − 2 a 2 + 5
= 7( a 4 8 ) − 2 a 2 + 5
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.
Let 2x + 1 = 0
∴ x = 1 2
4 = 4 x 3 + 3 x 2 + mx + 2 Let h(x) = − 4 − 4 = 4 (− 1 2 ) 3 + 3 (− 1 2 ) 2 + m(− 1 2 ) + 2 Let x = 1 2
4 = 4( 1 8 ) + 3( 1 4 ) − 1 2 m + 2
Simplify 6 = − 1 2 + 3 4 − 1 2 m Simplify − 6 = 1 4 − 1 2 m
Simplify
24 = 1 − 2m × LCM of 4
25 = − 2m
25 2 = m
12. If f(x) = x 2 − 4x + 5 is divided by x − p, the remainder is 2. Determine the value(s) of p.
Let x − p = 0
∴ x = p
2 = ( p) 2 − 4(p) + 5
0 = p 2 − 4p + 3
0 = (p − 3)(p − 1)
p = 3 OR p = 1
Let f(p) = 2, substitute p for x
Simplify
Factorise the trinomial
13. Determine the value of p if f(2) = 0 and it is given that f(x) = x 2 + 8p + 12.
( 2) 2 + 8p + 12 = 0
8p = − 16
Let f(2) = 0
Simplify p = − 2
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.
Let x − 1 = 0
∴ x = 1
f(1) = a( 1) 3 − b( 1) 2 + 4(1) + 5
11 = a − b + 4 + 5
Substitute 1 for x
Remainder is 11
a − b = 2 Simplify
a = b + 2 ① Simplify ①
Let 2x + 1 = 0 ∴ x = − 1 2
f( 1 2 ) = a ( 1 2 ) 3 − b ( 1 2 ) 2 + 4( 1 2 ) + 5 Substitute 1 2 for x
2 = − 1 8 a − 1 4 b − 2 + 5
Remainder is 2 and simplify
a − 2b = − 8 × LCM of 8
a = − 2b + 8 ② Simplify ② b + 2 = − 2b + 8 Equate ① to ②
3b = 6 Simplify ∴ b = 2
Substitute b = 2 back into ①: a = 2 + 2
∴ a = 4
a = 3 Sample
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. Let x + 3 = 0
x = − 3
44 = (− 3) 3 + a( − 3) + b
Remainder is 44, let x = − 3
44 = − 27 − 3a + b Simplify
17 = − 3a + b
b = 3a − 17 ①
Let x − 2 = 0
∴ x = 2
6 = (2) 3 + a(2) + b Remainder is 6, let x = 2
6 = 8 + 2a + b Simplify
2 = 2a + b
b = − 2a − 2 ②
∴ 3a − 17 = − 2a − 2 Let ① = ②
5a = 15
Simplify
Substitute a = 3 back into ①:
b = 3(3) − 17
b = − 8
2. FACTOR THEOREM
Substitute 3 for a
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.
Let x + 2 = 0
∴ x = − 2
q = ( 2) 3 + p( 2) 2 + 3 Remainder is q, substitute 2 for x
q = − 8 + 4p + 3 Simplify
q = − 5 + 4p
4p = q + 5
p = q + 5 4 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?
Let x − m = 0
∴ x = m
g(m) = 2( m) 3 + 7( m) 2 − m(m + 3)
Substitute m for x = 2 m 3 + 7 m 2 − m 2 − 3m Simplify = 2 m 3 + 6 m 2 − 3m
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.
Let x − 1 = 0 ∴ x = 1 ∴ 3( 1) 3 − m (1) 2 − 6(1) + m = 5 (1) 3 −
Sample
Equate expressions 3 − m − 6 + m = 5 − m − m Simplify 3 = 5 − 2m 8 = − 2m m = 4
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:
+ 1
= x 2(x − 1) − (x − 1)
= (x − 1)( x 2 − 1)
= (x − 1)(x − 1)(x + 1)
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
• Revision exercises to refresh prior knowledge.
• Detailed explanations of concepts and techniques.
• Worked examples help learners to better understand new concepts.
• Varied exercises to entrench theory and practise mathematical skills.
• Test papers and memorandums for exam preparation
• Formula sheets and accepted geometrical reasons for quick reference.
• Index of mathematical terms.
• The facilitator’s guide contains step-by-step calculations and answers.
• Use in school or at home.
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