Example and Explanation of Factoring Trinomals?

Algebra
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May 27th, 2015

FACTORING TRINOMIALS

2nd Level:

Positive leading term

Quadratics in different arguments

FLesson 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 thesecond 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

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:

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)

l)  5x2 − 22x + 21  = (5x − 7)(x − 3)

Example 1.   1 Lesson 16.)

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).
 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(x2 − x − 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).

[img src="http://www.themathpage.com/alg/alg_IMG/end.gif" width="372" height="40" alt="end" id="level" >

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.

abc 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 − 3b)  (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)

"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

http://www.themathpage.com/alg/factoring-trinomials.htm 

May 27th, 2015

Thank you so much!!! :D ur very smart!!! Xoxo

May 27th, 2015

no problem :) 

May 27th, 2015

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