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, x^{2}, y^{2}, z^{2}, 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) | = | abx^{2}_{ } | + | (ad | + | cb)x | + | cd |
showing that the coefficient of the square term, x^{2}, 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) | = | 2x^{2} + 2x + 3x + 3 |
= | 2x^{2} + 5x + 3 |
This is incorrect. The second choice is:
(2x + 1)(x + 3) | = | 2x^{2} + 6x + x + 3_{ } |
= | 2x^{2} + 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) | = | 10x^{2}_{ } + 4x + 5x + 2 |
= | 10x^{2}_{ } + 9x + 2 |
Exercise 1: Factorise each of the following expressions:
Solution: In this case we have:
2x^{2} + 5x + 3_{ } | = | (2x + 3)(x + 1) |
Solution: In this case we have:
3x^{2} + 7x + 2_{ } | = | (3x + 1)(x + 2) |
Solution: In this case we have:
3y^{2} − 5y − 2_{ } | = | (3y + 1)(y − 2) |
Solution: In this case we have:
4z^{2} − 23z + 15_{ } | = | (4z − 3)(z − 5) |
Solution: In this case we have:
64z^{2} + 4z − 3_{ } | = | (16z − 3)(4z + 1) |
Solution: This is a case of the difference of two squares (seen in the Brackets module).
4w^{2} − 25_{ } | = | (2w − 5)(2w + 5) |
Click on questions to reveal their solutions
Quiz 1: To which of the following does 12x^{2} + 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) | = | 12x^{2} − 7x + 24x − 14 |
= | 12x^{2} + 17x − 14 |
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:
x^{2} + 6x + 8 | = | (x + 2)(x + 4) |
The solution to the equation may now be obtained:
If x^{2} + 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 x^{2} − 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 2x^{2} + 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
3x^{2} + 7x + 2 | = | (3x + 1)(x + 2) |
so that if 3x^{2} + 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 3y^{2} − 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 4z^{2} − 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 64z^{2} + 4z − 3is thus z = 3 ⁄ 16 or z = −1 ⁄ 4.
Solution: In this case we have, from Exercise 1(f), that:
4w^{2}−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: 12x^{2} + 17x − 14 = 0?
Explanation: This quadratic is the one that occurs in Quiz 1. There, it was seen that 12x^{2} + 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.
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: x^{2} + 6x + 9 = (x + 3)^{2}
Solution: x^{2} + 4x + 4 = (x + 2)^{2}
Solution: x^{2} − 2x + 1 = (x − 1)^{2}
Solution: x^{2} − 2ax + a^{2} = (x − a)^{2}
The last example, x^{2} − 2ax + a^{2} = (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 x^{2} − 2ax + a^{2} with x^{2} − 10x + 25 we see that 2a = 10 and a^{2} = 25. Thus a = 5 is the solution. It may easily be checked that
(x − 5)^{2} = x^{2} − 10x + 25
Solution: Comparing z^{2} − 2az + a^{2} with z^{2} + 8z + 16 we see that −2a = 8 and a^{2} = 16. In this case, a = −4. Checking
(z − (−4))^{2} = (z + 4)^{2} = z^{2} + 8z + 16
Solution: Comparing w^{2} − 2aw + a^{2} with w^{2} − w + 1⁄4 we have that −2a = −1 and a^{2} = 1⁄4. In this case, a = 1⁄2. It is now easy to check that
(w − 1⁄2)^{2} = w^{2} − w + 1⁄4
Solution: Comparing y^{2} − 2ay + a^{2} with y^{2} + 5y + 25⁄4 we have that −2a = 5 and a^{2} = 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} = y^{2} + 5y + 25 ⁄ 4
Click on questions to reveal their solutions
Quiz 3: Which of the following quadratic expressions is a complete square?
Explanation: Comparing z^{2} − 2az + a^{2} with z^{2} − 3z + 9⁄4 we have that −2a = −3 and a^{2} = 9⁄4. Thus a = 3⁄2. It is now easy to check that
(z − 3⁄2)^{2} = z^{2} − 3y + 9 ⁄ 4
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 x^{2} − 4x + 5. From the beginning of Section 3 we note that
x^{2} − 4x + 4 = (x − 2)^{2}.
Since
x^{2} − 4x + 5 = (x^{2} − 4x + 4) + 1
the expression may be written as
x^{2} − 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 x^{2} + 6x + 11 = (x^{2} + 6x + 9) + 2, the expression may be written x^{2} + 6x + 11 = (x + 3)^{2} + 2.
Solution: Here we note that x^{2} + 4x + 3 = (x^{2} + 4x + 4) − 1. Now we may use the fact that (x + 2)^{2} = (x^{2} + 4x + 4) to obtain x^{2} + 4x + 3 = (x + 2)^{2} − 1.
Solution: Since 2x^{2} + 8x + 4 = 2(x^{2} + 4x + 2) = 2[(x^{2} + 4x + 4) − 2], we have 2x^{2} + 8x + 4 = 2[(x + 2)^{2} − 2] = 2(x + 2)^{2} − 4.
Solution: Here x^{2} − 2ax + a^{2} + b^{2} = (x − a)^{2} + b^{2}.
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 x^{2} + 6x + 9 and this information was used to complete the square on x^{2} + 6x + 10. In practice the starting point is usually the quadratic expression, i.e., in (a) it would be x^{2} + 6x + 10. The problem is to work directly from this. The general procedure uses the following observation:
so the expression x^{2} + 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:
x^{2} + 8x + 15 | = | [x^{2} + 8x] + 15 |
= | [x^{2} + 8x + (8 ⁄2)^{2} − (8 ⁄2)^{2}] + 15 | |
= | [(x + 4)^{2} − 4^{2}] + 15 | |
= | (x + 4)^{2} − 1 |
Solution: For x^{2} − 5x + 6 the procedure is much the same:
x^{2} − 5x + 6 | = | [x^{2} − 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
x^{2} − 6x + 5 | = | [x^{2} − 6x] + 5 |
= | [x^{2} − 6x + (3)^{2} − (3)^{2}] + 5 | |
= | [(x − 3)^{2} − 9] + 5 | |
= | (x − 3)^{2} − 4 |
Solution:
2z^{2} + 8z + 9 | = | 2[z^{2} + 4z] + 9 |
= | 2[z^{2} + 4z + (2)^{2} − (2)^{2}] + 9 | |
= | 2[(z + 2)^{2} − 4] + 9 | |
= | 2(z + 2)^{2} + 1 |
Solution:
2w^{2} − 5w + 7 | = | 2[w^{2} − 5 ⁄ 2 w] + 7 |
= | 2[w^{2} − 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:
3y^{2} + 2y + 2 | = | 3[y^{2} + 2⁄3 y] + 2 |
= | 3[y^{2} + 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 2x^{2} − 3x + 5?
Explanation: We have
2x^{2} − 3x + 5 | = | 2[x^{2} − 3 ⁄ 2 x] + 5 |
= | 2[x^{2} − 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 |
In each of the following, choose: