ABSTRACT
The zero’s shown in the dividend are not normally shown, but are included to clarify the subtraction process and to keep similar terms in their respective columns.
Problem 4. Divide (x2 + 3x − 2) by (x − 2)
x − 2 x + 5) x2 + 3x − 2 x2 − 2x ______
5x − 2 5x − 10 ______
8 ______
Hence x2 + 3x − 2
x − 2 = x+ 5+ 8
x− 2 Problem 5. Divide 4a3 − 6a2b + 5b3 by 2a − b
2a − b 2a2 − 2ab − b2)
4a3 − 6a2b + 5b3 4a3 − 2a2b _________
−4a2b + 5b3 −4a2b + 2ab2 ____________
−2ab2 + 5b3 −2ab2 + b3 ___________
4b3 ___________
Thus
4a3 − 6a2b + 5b3 2a − b = 2a
2− 2ab− b2+ 4b 3
2a− b
Now try the following Practice Exercise
Practice Exercise 25 Polynomial division (Answers on page 659)
1. Divide (2x2 + xy − y2) by (x + y) 2. Divide (3x2 + 5x − 2) by (x + 2) 3. Determine (10x2 + 11x − 6)÷ (2x + 3)
4. Find: 14x2 − 19x − 3
2x − 3 5. Divide (x3 + 3x2y + 3xy2 + y3) by (x + y) 6. Find (5x2 − x + 4)÷ (x − 1) 7. Divide (3x3 + 2x2 − 5x + 4) by (x + 2)
8. Determine: 5x4 + 3x3 − 2x + 1
x − 3
There is a simple relationship between the factors of a quadratic expression and the roots of the equation obtained by equating the expression to zero. For example, consider the quadratic equation x2 + 2x − 8 = 0 To solve this we may factorise the quadratic expression x2 + 2x − 8 giving (x − 2)(x + 4) Hence (x − 2)(x + 4) = 0 Then, if the product of two number is zero, one or both of those numbers must equal zero. Therefore,
either (x − 2) = 0, from which, x = 2 or (x + 4) = 0, from which, x = −4 It is clear then that a factor of (x − 2) indicates a root of +2, while a factor of (x + 4) indicates a root of −4. In general, we can therefore say that:
a factor of (x− a) corresponds to a root of x= a In practice, we always deduce the roots of a simple quadratic equation from the factors of the quadratic expression, as in the above example. However, we could reverse this process. If, by trial and error, we could determine that x = 2 is a root of the equation x2 + 2x − 8 = 0 we could deduce at once that (x − 2) is a factor of the expression x2 + 2x − 8. We wouldn’t normally solve quadratic equations this way — but suppose we have to factorise a cubic expression (i.e. one in which the highest power of the variable is 3). A cubic equation might have three simple linear factors and the difficulty of discovering all these factors by trial and error would be considerable. It is to deal with this kind of case that we use the factor theorem. This is just a generalised version of what we established above for the quadratic expression. The factor theorem provides a method of factorising any polynomial, f (x), which has simple factors. A statement of the factor theorem says:
‘if x= a is a root of the equation f (x)= 0, then (x− a) is a factor of f (x)’
The following worked problems show the use of the factor theorem.
