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17

FACTORING TRINOMIALS

2nd Level:

Positive leading term

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 signsexclamation

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 (that is, 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 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 termexclamation

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)  2x2x − 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)  2x2x − 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)

l)  5x2 − 22x + 21  = (5x − 7)(x − 3)

Example 1.   1 the coefficient of x2.   Factor  x2 + 3x − 10.

Solution.   The binomial factors will look like this:

(x   a)(x   b)

Since the coefficient of x2 is 1, it will not matter in which binomial we put the numbers.

(x   a)(x   b) = (x   b)(x   a).

Now, what are the factors of 10?  Let us try 2 and 5:

x2 + 3x − 10 = (x   2)(x   5).

We must now choose the signs so that -- as always -- the sum of the outers plus the inners will equal the middle term, which is +3x.

Choose +5 and −2.

x2 + 3x − 10 = (x2)(x + 5).

When the coefficient of x2 is 1, we may simply look for two numbers whose product is 10, the numerical term, and whose algebraic sum is +3, the coefficient of x.  That will ensure that the sum of the outers plus the inners will equal the middle term.

Again, the order of the factors does not matter.

(x − 2)(x + 5) = (x + 5) (x − 2).

Example 2.   Factor  x2x − 12.

   Solution. (x   ?)(x   ?)

There are several factors of 12.  Let us try 2 and 6.

(x   2)(x   6)

But there is no way to choose signs for 6x and 2x to give the middle term, which is −x.

Let's try 3 and 4:

(x   3)(x   4)

That will work. Choose −4 and +3.

(x + 3)(x − 4) = x2x − 12.

Skill in factoring depends on skill in multiplying, specifically in constructing the middle term. (Lesson 16.)

Problem 4.   Factor.  Again, the order of the factors does not matter.

a)  x2 + 5x + 6  = (x + 2)(x + 3)

b)  x2x − 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)  x2x − 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).
  Now 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(x2x − 6) = 4(x + 2)(x − 3)

e)  6x3 + 10x2 − 4x  = 2x(3x2 + 5x − 2) = 2x(3x − 1)(x + 2)

f)  12x10 + 42x9 + 18x8  = 6x8(2x2 + 7x + 3) = 6x8(2x + 1)(x + 3).

end

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  = −(x2x − 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 in this way:

3x2 + 2x − 1 = (3x − 1)(x + 1),

then the quadratic with argument x3is factored 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)  x4x2 − 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 that example you are doing trigonometry, you are doing nothing but algebra.

end

Next Lesson:  Perfect Square Trinomials

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