17
FACTORING TRINOMIALS
Quadratics in different arguments
FACTORING IS THE REVERSE of multiplying. Skill in factoring, then, depends upon skill in multiplying (Lesson 16). As for a quadratic trinomial --
2x² + 9x − 5
-- it will be factored as a product of binomials:
(? ?)(? ?)
Now, how will 2x² be produced? There is only one way: 2x· x :
(2x ?)(x ?)
And how will 5 be produced? Again, there is only one way: 1· 5. But does the 5 go with 2x or with x ?
(2x 5)(x 1) or (2x 1)(x 5) ?
Notice: We have not yet placed any signs!
How shall we decide between these two possibilities? It is the combination that will correctly give the middle term, 9x :
2x² + 9x − 5.
Consider the first possibility:
(2x 5)(x 1)
Is it possible to produce 9x by combining the outers and the inners: 2x· 1 with 5x ?
No, it is not. Therefore, we must eliminate that possibility and consider the other:
(2x 1)(x 5)
Can we produce 9x by combining 10x with x ?
Yes -- if we choose +5 and −1:
(2x − 1)(x + 5)
(2x − 1)(x + 5) = 2x² + 9x − 5
Skill in factoring depends on skill in multiplying -- particularly in picking out the middle term!
Problem 1. Place the correct signs to give the middle term.
a) 2x² + 7x − 15 = (2x − 3)(x + 5)
b) 2x² − 7x − 15 = (2x + 3)(x − 5)
c) 2x² − x − 15 = (2x + 5)(x − 3)
d) 2x² − 13x + 15 = (2x − 3)(x − 5)
Note: When the constant term is negative, as in parts a), b), c), then the signs will be different. But when that term is positive, as in part d), the signs will be the same. Usually, however, that happens by itself.
Nevertheless, can you correctly factor the following?
2x² − 5x + 3 = (2x − 3)(x − 1)
Problem 2. Factor these trinomials.
a) 3x² + 8x + 5 = (3x + 5)(x + 1)
b) 3x² + 16x + 5 = (3x + 1)(x + 5)
c) 2x² + 9x + 7 = (2x + 7)(x + 1)
d) 2x² + 15x + 7 = (2x + 1)(x + 7)
e) 5x² + 8x + 3 = (5x + 3)(x + 1)
f) 5x² + 16x + 3 = (5x + 1)(x + 3)
Problem 3. Factor these trinomials.
a) 2x² − 7x + 5 = (2x − 5)(x − 1)
b) 2x² − 11x + 5 = (2x − 1)(x − 5)
c) 3x² + x − 10 = (3x − 5)(x + 2 )
d) 2x² − x − 3 = (2x − 3)(x + 1)
e) 5x² − 13x + 6 = (5x − 3)(x − 2)
f) 5x² − 17x + 6 = (5x − 2)(x − 3)
g) 2x² + 5x − 3 = (2x − 1)(x + 3)
h) 2x² − 5x − 3 = (2x + 1)(x − 3)
i) 2x² + x − 3 = (2x + 3)(x − 1)
j) 2x² − 13x + 21 = (2x − 7 )(x −3)
k) 5x² − 7x − 6 = (5x + 3)(x − 2)
i) 5x² − 22x + 21 = (5x − 7)(x − 3)
Example 1. 1 the coefficient of x². Factor x² + 3x − 10.
Solution. The binomial factors will have this form:
(x a)(x b)
What are the factors of 10? Let us hope that they are 2 and 5:
x² + 3x − 10 = (x 2)(x 5)
We must now choose the signs so that the coefficient of the middle term will be +3. Choose −2 and + 5.
x² + 3x − 10 = (x − 2)(x + 5)
Note: When 1 is the coefficient of x², the order of the factors does not matter.
x² + 3x − 10 = (x + 5) (x − 2)
Example 2. Factor x² − x − 12.
Solution. We must find factors of 12 whose algebraic sum will be the coefficient of x : −1. Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3)
Problem 4. Factor. Again, the order of the factors does not matter.
a) x² + 5x + 6 = (x + 2)(x + 3)
b) x² − x − 6 = (x − 3 )(x + 2)
c) x² + x − 6 = (x + 3 )(x − 2)
d) x² − 5x + 6 = (x − 3)(x − 2 )
e) x² + 7x + 6 = (x + 1)(x + 6 )
f) x² − 7x + 6 = (x − 1)(x − 6 )
g) x² + 5x − 6 = (x − 1)(x + 6 )
h) x² − 5x − 6 = (x + 1)(x − 6 )
Problem 5. Factor.
a) x² − 10x + 9 = (x − 1 )(x − 9)
b) x² + x − 12 = (x + 4)(x − 3)
c) x² − 6x − 16 = (x − 8)(x + 2)
d) x² − 5x − 14 = (x − 7)(x + 2)
e) x² − x − 2 = (x + 1)(x − 2)
f) x² − 12x + 20 = (x − 10 )(x − 2)
g) x² − 14x + 24 = (x − 12 )(x − 2)
Example 3. Factor completely 6x8 + 30x7 + 36x6.
Solution. To factor completely means to first remove any common factors (Lesson 15).
| 6x8 + 30x7 + 36x6 | = | 6x6(x² + 5x + 6) |
| Continue by factoring the trinomial: | ||
| = | 6x6(x + 2)(x + 3) | |
Problem 6. Factor completely. First remove any common factors.
a) x3 + 6x² + 5x = x(x2 + 6x + 5) = x(x + 5)(x + 1)
b) x5 + 4x4 + 3x3 = x3(x2 + 4x + 3) = x3(x + 1)(x + 3)
c) x4 + x3 − 6x² = x²(x² + x − 6) = x²(x + 3)(x − 2)
d) 4x² − 4x − 24 = 4(x² − x − 6) = 4(x + 2)(x − 3)
e) 2x3 − 14x² − 36x = 2x(x2 − 7x − 18) = 2x(x + 2)(x − 9)
f) 12x10 + 42x9 + 18x8 = 6x8(2x² + 7x + 3) = 6x8(2x + 1)(x + 3)
2nd Level
Example 4. Factor by making the leading term positive.
−x² + 5x − 6 = −(x² − 5x + 6) = −(x − 2)(x − 3).
Problem 7. Factor by making the leading term positive.
a) −x² − 2x + 3 = −(x² + 2x − 3) = −(x + 3)(x − 1)
b) −x² + x + 6 = −(x² − x − 6) = −(x + 2)(x − 3)
c) −2x² − 5x + 3 = −(2x² + 5x − 3) = −(2x − 1)(x + 3)
Quadratics in different arguments
Here is the form of a quadratic trinomial with argument x :
ax² + bx + c.
The argument is whatever is being squared. x is being squared. x is called the argument. The argument appears in the middle term.
a, b, c are called constants. In this quadratic,
3x² + 2x − 1,
the constants are 3, 2, −1.
Now here is a quadratic whose argument is x3:
3x6 + 2x3 − 1.
x6 is the square of x3. (Lesson 13: Exponents.)
But that quadratic has the same constants -- 3, 2, − 1 -- as the one above. In a sense, it is the same quadratic only with a different argument. For it is the constants that distinguish a quadratic.
Now, since the quadratic with argument x can be factored as
3x² + 2x − 1 = (3x − 1)(x + 1),
then the quadratic with argument x3 is factored in the same way:
3x6 + 2x3 − 1 = (3x3 − 1)(x3 + 1).
Whenever a quadratic has constants 3, 2, −1, then for any argument, the factoring will be
(3 times the argument − 1)(argument + 1).
| Example 5. | z² − 3z − 10 | = | (z + 2)(z − 5). | |
| x8 − 3x4 − 10 | = | (x4 + 2)(x4 − 5). | ||
The trinomials on the left have the same constants 1, −3, −10 but different arguments. That is the only difference between them. In the first, the argument is z. In the second, the argument is x4.
(The square of x4 is x8.)
Each quadratic is factored as
(argument + 2)(argument − 5).
Every quadratic with constants 1, −3, −10 will be factored that way.
Problem 8.
a) Write the form of a quadratic trinomial with argument z.
az² + bz + c
b) Write the form of a quadratic trinomial with argument x4.
ax8 + bx4 + c
c) Write the form of a quadratic trinomial with argument xn.
ax2n + bxn + c
Problem 10. Multiply out each of the following, which have the same constants, but different argument.
| a) | (z + 3)(z − 1) = z² + 2z − 3 | b) | (y + 3)(y − 1) = y² + 2y − 3 |
c) (y6 + 3)(y6 − 1) = y12 + 2y6 − 3
d) (x5 + 3)(x5 − 1) = x10 + 2x5 − 3
Problem 11. Factor each quadratic.
a) x² − 6x + 5 = (x − 1)(x − 5)
b) z² − 6z + 5 = (z − 1)(z − 5)
c) x8 − 6x4 + 5 = (x4 − 1)(x4 − 5)
d) x10 − 6x5 + 5 = (x5 − 1)(x5 − 5)
e) x6y6 − 6x3y3 + 5 = (x3y3 − 1)(x3y3 − 5)
| f) sin²x − 6 sin x + 5 | = | (sin x − 1)(sin x − 5). |
| sin²x -- "sine squared x" -- means (sin x)². | ||
Problem 12. Factor each quadratic.
a) x4 − x² − 2 = (x² − 2)(x² + 1)
b) y6 + 2y3 − 8 = (y3 + 4)(y3 − 2)
c) z8 + 4z4 + 3 = (z4 + 1)(z4 + 3)
d) 2x10 + 5x5 + 3 = (2x5 + 3)(x5 + 1)
e) x4y² − 3x²y − 10 = (x²y + 2)(x²y − 5)
f) cos²x − 5 cos x + 6 = (cos x − 3)(cox x − 2)
Next Lesson: Perfect Square Trinomials
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