17

FACTORING TRINOMIALS

2nd Level:

Positive leading term

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:

(?   ?)(?   ?)

The first term of each binomial will be the factors of 2x², and the second term will be the factors of 5.  Now, how can we produce 2x²? There is only one way:  2x· x :

(2x   ?)(x   ?)

And how can we produce 5?  Again, there is only one way:  1·  5.  But does the 5 go with  2x --

(2x   5)(x   1)

or with  x --

(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 :

Is it possible to produce  9x  by combining the outers and the inners:  2x· 1 with 5· x ?

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 1x ?

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 in each factor must be different.  But when the constant term is positive, as in part d), the signs must 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)

Solution.   To factor completely means to first remove any common factors (Lesson 15).

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).

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)

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 in this way:

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).

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.

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)

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)

"cos x" is an abbreviation for the trigonometric function "cosine of angle x." It is conventional to write the square of cos x as cos²x ("cosine squared x"). In calculus, rather than solve triangles with cos x, we do algebra with it. The above is an example. And so while you may think that in this example you are doing trigonometry, you are doing nothing but algebra

Next Lesson:  Perfect Square Trinomials

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