1. Introduction
In the module on Factorising Expressions we looked at how to factorise quadratic expressions which have the number 1 in front of the highest order term, x2, y2, z2, etc.. If the highest order term has a number other than this then more work must be done to factorise the expression. As in the earlier case, some insight is gained by looking at a general expression with factors (ax + c) and (bx + d). Then:
| F | O | I | L | |||||
| (ax + c)(bx + d) | = | axbx | + | axd | + | cbx | + | cd |
| (ax + c)(bx + d) | = | abx2 | + | (ad | + | cb)x | + | cd |
showing that the coefficient of the square term, x2, is ab, the product of the coefficients of the x-terms in each factor. The coefficient of the x-term is made up from the coefficients as follows:
This is the information needed to find the factors of quadratic expressions.
Example 1: Factorise the following expressions.
Solution:
The factors of 2 are 2 and 1, and the factors of 3 are 3 and 1. If the quadratic expression factorises then it is likely to be of the form (2x + c)(1x + d) and the choice for c, d is 3, 1 or 1, 3. Trying the first combination:| (2x + 3)(x + 1) | = | 2x2 + 2x + 3x + 3 |
| = | 2x2 + 5x + 3 |
This is incorrect. The second choice is:
| (2x + 1)(x + 3) | = | 2x2 + 6x + x + 3 |
| = | 2x2 + 7x + 3 |
This is the correct factorisation.
Solutions: There is more than one choice for the first term since 10 is 1 × 10 as well as 2 × 5. The final term will factor as 2 × 1. Which combination of pairs, either (1, 10) with (2, 1), or (2, 5) with (2, 1), will give the correct coefficient of x, i.e., 9? The latter two pairs seem the more likely since 2 × 2 + 5 × 1 = 9. Checking this:
| (2x + 1)(5x + 2) | = | 10x2 + 4x + 5x + 2 |
| = | 10x2 + 9x + 2 |
Exercise 1: Factorise each of the following expressions:
Solution: In this case we have:
| 2x2 + 5x + 3 | = | (2x + 3)(x + 1) |
Solution: In this case we have:
| 3x2 + 7x + 2 | = | (3x + 1)(x + 2) |
Solution: In this case we have:
| 3y2 − 5y − 2 | = | (3y + 1)(y − 2) |
Solution: In this case we have:
| 4z2 − 23z + 15 | = | (4z − 3)(z − 5) |
Solution: In this case we have:
| 64z2 + 4z − 3 | = | (16z − 3)(4z + 1) |
Solution: This is a case of the difference of two squares (seen in the Brackets module).
| 4w2 − 25 | = | (2w − 5)(2w + 5) |
Click on questions to reveal their solutions
Quiz 1: To which of the following does 12x2 + 17x − 14 factorise?
Explanation: There are several possibilities, since the final term is −14 and the two quantities corresponding to c and d must therefore have opposite signs. The possible factors of 12 are (1, 12), (2, 6), (3, 4). For −14, the possible factors are (±1, ±14), (±2, ±7). It is now a matter of trial and error. The possible combinations are:
| (1, 12) | and | (±1, ±14), | (1, 12) | and | (±2, ±7), |
| (2, 6) | and | (±1, ±14), | (2, 6) | and | (±2, ±7), |
| (3, 4) | and | (±1, ±14), | (3, 4) | and | (±2, ±7), |
By inspection, (2 × 12) + (1 × (−7)) = 24 − 7 = 17, so the factors appear to be (x + 2) and (12x − 7). This can easily be checked:
| (x + 2)(12x + 7) | = | 12x2 − 7x + 24x − 14 |
| = | 12x2 + 17x − 14 |
2. Solving Quadratic Equations
Factorising a quadratic expression and finding the roots of a quadratic equation are closely related.
Example 2: Find the solutions to the equations:
Solution: The quadratic expression will factorise as follows:
| x2 + 6x + 8 | = | (x + 2)(x + 4) |
The solution to the equation may now be obtained:
| If x2 + 6x + 8 | = | (x + 2)(x + 4) |
| then (x + 2)(x + 4) | = | 0 |
Thus, either (x + 2) = 0, or (x + 4) = 0. The solution to the equation is thus x = −2, or x = −4
Solution: In this example the expression is:
The solution to the equation x2 − 4x + 4 is thus x = 2. In this case, the equation is said to have equal roots.
Exercise 2: Find the solution to each of the following equations.
Solution: In this case we have, from Exercise 1(a), that
Thus if 2x2 + 5x + 3 = 0 then we have
either (2x + 3) = 0, or (x + 1) = 0.
For the first of these:
| 2x + 3 | = | 0 | |
| 2x | = | −3 | (adding −3 to both sides) |
| x | = | −3 ⁄ 2 | (dividing both sides by 2) |
The solution to the second is obviously x=−1
Solution: In this case we have, from Exercise 1(b), that
| 3x2 + 7x + 2 | = | (3x + 1)(x + 2) |
| so that if 3x2 + 7x + 2 | = | 0 |
| then (3x + 1)(x + 2) | = | 0 |
Thus either (3x + 1) = 0 or (x + 2) = 0
For the first of these:
| 3x + 1 | = | 0 |
| 3x | = | −1 (adding −1 to both sides) |
| x | = | −1⁄3 (dividing both sides by 3) |
The solution to the second part is obviously x=−2. The solution to the original equation is thus x = −2 or x = −1 ⁄ 3.
Solution: In this case we have, from Exercise 1(c), that:
Thus, either 3y + 1 = 0, or y−2 = 0. For the first part:
| 3y + 1 | = | 0 |
| 3y | = | −1 (adding −1 to both sides) |
| y | = | −1 ⁄ 3 (dividing both sides by 3) |
The solution to the second part is obviously y = 2. The quadratic equation 3y2 − 5y − 2 = 0 thus has the solution y = −1 ⁄ 3 or y = 2.
Solution: In this case we have, from Exercise 1(d), that:
Thus, either 4z−3 = 0 or z−5 = 0.
Proceeding as in the previous parts of this exercise, the solution to the first part is z = 3⁄4 and to the second part is z = 5.
The solution to 4z2 − 23z + 15 = 0 is therefore z = 3⁄4 or z = 5.
Solution: In this case we have, from Exercise 1(e), that
Thus either (16z − 3) = 0 or (4z + 1) = 0. For the first part:
| 16z − 3 | = | 0 |
| 16z | = | 3 (adding 3 to both sides) |
| z | = | 3⁄16 (dividing both sides by 16) |
For the second part,
| 4z + 1 | = | 0 |
| 16z | = | −1 (adding −1 to both sides) |
| z | = | −1⁄4 (dividing both sides by 4) |
The solution to the equation 64z2 + 4z − 3is thus z = 3 ⁄ 16 or z = −1 ⁄ 4.
Solution: In this case we have, from Exercise 1(f), that:
4w2−25 = (2w − 5)(2w + 5) = 0
The solution to this is:
i.e.w = ±5 ⁄ 2
Click on questions to reveal their solutions
Quiz 2: Which of the following is the solution to the quadratic equation: 12x2 + 17x − 14 = 0?
Explanation: This quadratic is the one that occurs in Quiz 1. There, it was seen that 12x2 + 17x − 14 = (x + 2)(12x − 7), so either x + 2 = 0 or 12x − 7 = 0.
The solution to the first is x = − 2, and to the second is x = 7 ⁄ 12.
3. Complete Squares
In Example 2(b) we encountered the quadratic expression:
This expression is called a complete square. Quadratic expressions which can be factored into a complete square are useful in many situations. They have a particularly simple structure and it is important to be able to recognise such factorisations.
Example 3: Show that the following quadratic expressions are complete squares:
Solution: x2 + 6x + 9 = (x + 3)2
Solution: x2 + 4x + 4 = (x + 2)2
Solution: x2 − 2x + 1 = (x − 1)2
Solution: x2 − 2ax + a2 = (x − a)2
The last example, x2 − 2ax + a2 = (x − a)2, is a general case and it may be used to find perfect squares for any given example. It may be usefully employed in finding solutions to the following exercises:
Exercise 3: Write each of the following as a complete square.
Solution: Comparing x2 − 2ax + a2 with x2 − 10x + 25 we see that 2a = 10 and a2 = 25. Thus a = 5 is the solution. It may easily be checked that
(x − 5)2 = x2 − 10x + 25
Solution: Comparing z2 − 2az + a2 with z2 + 8z + 16 we see that −2a = 8 and a2 = 16. In this case, a = −4. Checking
(z − (−4))2 = (z + 4)2 = z2 + 8z + 16
Solution: Comparing w2 − 2aw + a2 with w2 − w + 1⁄4 we have that −2a = −1 and a2 = 1⁄4. In this case, a = 1⁄2. It is now easy to check that
(w − 1⁄2)2 = w2 − w + 1⁄4
Solution: Comparing y2 − 2ay + a2 with y2 + 5y + 25⁄4 we have that −2a = 5 and a2 = 25⁄4. From this, it must follow that a = 5⁄2, as (−5⁄2)2 = (−5)2⁄(2)2 = 25⁄4. It is now easy to check that
(y − (−5⁄2))2 = (y + 5⁄2)2 = y2 + 5y + 25 ⁄ 4
Click on questions to reveal their solutions
Quiz 3: Which of the following quadratic expressions is a complete square?
Explanation: Comparing z2 − 2az + a2 with z2 − 3z + 9⁄4 we have that −2a = −3 and a2 = 9⁄4. Thus a = 3⁄2. It is now easy to check that
(z − 3⁄2)2 = z2 − 3y + 9 ⁄ 4
4. Completing the Square
Not every quadratic is a complete square but it is possible to write all quadratics as a complete square plus a number. This is the process known as completing the square. This is an extremely useful algebraic procedure with many applications.
As an example, let us take the quadratic expression x2 − 4x + 5. From the beginning of Section 3 we note that
x2 − 4x + 4 = (x − 2)2.
Since
x2 − 4x + 5 = (x2 − 4x + 4) + 1
the expression may be written as
x2 − 4x + 5 = (x − 2)2 + 1
and we have completed the square on this quadratic.
Example 4: Use the results of Example 3 to complete the square on each of the following quadratic expressions.
Solution: Since x2 + 6x + 11 = (x2 + 6x + 9) + 2, the expression may be written x2 + 6x + 11 = (x + 3)2 + 2.
Solution: Here we note that x2 + 4x + 3 = (x2 + 4x + 4) − 1. Now we may use the fact that (x + 2)2 = (x2 + 4x + 4) to obtain x2 + 4x + 3 = (x + 2)2 − 1.
Solution: Since 2x2 + 8x + 4 = 2(x2 + 4x + 2) = 2[(x2 + 4x + 4) − 2], we have 2x2 + 8x + 4 = 2[(x + 2)2 − 2] = 2(x + 2)2 − 4.
Solution: Here x2 − 2ax + a2 + b2 = (x − a)2 + b2.
The starting position for each of the above cases was the correct choice of the complete square. For example, part (a) began with the expansion of (x + 3)2 as x2 + 6x + 9 and this information was used to complete the square on x2 + 6x + 10. In practice the starting point is usually the quadratic expression, i.e., in (a) it would be x2 + 6x + 10. The problem is to work directly from this. The general procedure uses the following observation:
so the expression x2 + px can be made into a complete square by adding (p ⁄ 2)2, i.e., by adding the square of half the coefficient of x. So as not to alter the expression the same amount must be subtracted. In other words we use the equality
and this is the essence of completing the square.
Example 5: Complete the square on the following expressions:
Solution: Completing the square means adding the square of half the coefficient of x and then subtracting the same amount. Thus:
| x2 + 8x + 15 | = | [x2 + 8x] + 15 |
| = | [x2 + 8x + (8 ⁄2)2 − (8 ⁄2)2] + 15 | |
| = | [(x + 4)2 − 42] + 15 | |
| = | (x + 4)2 − 1 |
Solution: For x2 − 5x + 6 the procedure is much the same:
| x2 − 5x + 6 | = | [x2 − 5x] + 6 |
| = | [(x − 5 ⁄ 2)2 − (5 ⁄ 2)2] + 6 | |
| = | [(x − 5 ⁄ 2)2 − 25 ⁄ 4] + 6 | |
| = | (x − 5 ⁄ 2)2 − 1 ⁄ 4 |
Exercise 4: Complete the square on each of the following.
Solution: Here
| x2 − 6x + 5 | = | [x2 − 6x] + 5 |
| = | [x2 − 6x + (3)2 − (3)2] + 5 | |
| = | [(x − 3)2 − 9] + 5 | |
| = | (x − 3)2 − 4 |
Solution:
| 2z2 + 8z + 9 | = | 2[z2 + 4z] + 9 |
| = | 2[z2 + 4z + (2)2 − (2)2] + 9 | |
| = | 2[(z + 2)2 − 4] + 9 | |
| = | 2(z + 2)2 + 1 |
Solution:
| 2w2 − 5w + 7 | = | 2[w2 − 5 ⁄ 2 w] + 7 |
| = | 2[w2 − 5⁄2 w + (5 ⁄ 4)2 − (5 ⁄ 4)2] + 7 | |
| = | 2[(w − 5 ⁄ 4)2 − 25 ⁄ 16] + 7 | |
| = | [2(w − 5 ⁄ 4)2 − 25 ⁄ 8] + 56 ⁄ 8 | |
| = | 2(w − 5 ⁄ 4)2 + 31 ⁄ 8 |
Solution:
| 3y2 + 2y + 2 | = | 3[y2 + 2⁄3 y] + 2 |
| = | 3[y2 + 2⁄3 y + (1 ⁄ 3)2 − (1 ⁄ 3)2] + 2 | |
| = | 3[(y + 1 ⁄ 3)2 − 1 ⁄ 9] + 2 | |
| = | [3(y + 1 ⁄ 3)2 − 1 ⁄ 3] + 6 ⁄ 3 | |
| = | 3(y + 1 ⁄ 3)2 + 5 ⁄ 3 |
Click on questions to reveal their solutions
Quiz 4: Which of the expressions below is obtained after completing the square on 2x2 − 3x + 5?
Explanation: We have
| 2x2 − 3x + 5 | = | 2[x2 − 3 ⁄ 2 x] + 5 |
| = | 2[x2 − 3 ⁄ 2 x + (3 ⁄ 4)2 − (3 ⁄ 4)2] + 5 | |
| = | 2[(x + 3 ⁄ 4)2 − 9 ⁄ 16] + 5 | |
| = | 2(x − 3 ⁄ 4)2 + 31 ⁄ 8 |
5. Quiz on Quadratics
In each of the following, choose: