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Common factor:  2nd Level

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The leading term

Proof of the Rule of Signs

Factoring by grouping

The unknown a common factor

In a polynomial, the leading term is the term with the highest exponent. Normally, it is the first term on the left.  In this polynomial,

3x2 − 6x + 9,

3x2 is the leading term.

Now we like the leading term to be positive.  (We will see that when we factor trinomials.)  Therefore, if we have the following,

−3x2 + 6x − 9,

we can make the leading term positive  by writing

−3x2 + 6x − 9  =  −3(x2 − 2x + 3).

We can remove a negative factor.

Problem 10.   Make the leading term positive.

   a)   −3x − 6 = −3(x + 2)   b)   −5x2 + 5x = −5x(x − 1)
 
   c)   x5x3 = −x3(x2 + 1)   d)   −2x2 + 6x − 2 = −2(x2 − 3x + 1)

e)  −32x3 − 12x2 + 8x  = −4x(8x2 + 3x − 2)

f)  −9x5 + 30x4 − 3x3  = −3x3(3x2 − 10x + 1)

Problem 11.   In each sum, remove the factor  xn  by displaying it on the left.

For example, xn + xn + 2 = xn(1 + x2).

   a)   xn + xn + 3 = xn(1 + x3)   b)   xn + xn + 4 = xn(1 + x4)
 
   c)   xn + 3xn = xn(x3 − 1)   d)   xn + 1 + xn = xn(x + 1)

Problem 12.   In each sum, factor out the lower power of x.

For example,  xn − 1 + xn = xn − 1(1 + x),

where xn − 1 is the lower power.  On multiplying out, we would add the exponents (Lesson 13) and obtain the left-hand side.

   a)   xn + 4 + xn + 1 = xn + 1(x3 + 1)   b)   xn + xn + 3 = xn(1 + x3)
 
   c)   xnxn − 2 = xn − 2(x2 − 1)   d)   xn − 1xn + 1 = xn − 1(1 − x2)

Problem 13.  The Rule of Signs.   By applying the definition of the negative of a number, prove that, if ab is positive, then  (−a)b  is the negative of ab.  That is, prove:  Unlike signs produce a negative number:

(−a)b = −ab.

ab + (−a)b = [a + (−a)]b = 0· b = 0.

Therefore, since a number has one and only one negative,

(−a)b = −ab.

Example 4.   x(x + 5) + 3(x + 5).

What is the common factor?

(x + 5) is the common factor.  Therefore,

x(x + 5) + 3(x + 5) = (x + 3)(x + 5)

This is similar to adding like terms.  In the first term, x is the coefficient of (x + 5).  In the second term, 3 is its coefficient.  We add the coefficients of (x + 5).  And we preserve the common factor on the right.

Problem 14.   Add the common factors. Do not remove parentheses.

a)   x(x + 1) + 2(x + 1)  = (x + 2)(x + 1)

b)  x(x − 2) − 3(x − 2)  = (x − 3)(x − 2)

c)  x(x + 1) − (x + 1)  = (x − 1)(x + 1)

d)  x2(x − 5) + 4(x − 5)  = (x2 + 4)(x − 5)

Example 5.  Factoring by grouping.   Factor  x3 −5x2 + 3x − 15.

Solution.   Group the first and second terms -- find their common factor. Do the same with the third and fourth terms.

x3 − 5x2 + 3x − 15 = x2(x − 5) + 3(x − 5)
 
  = (x2 + 3)(x − 5).

Problem 15.   Factor by grouping.

  a)  x3 + x2 + 3x + 3   =   x2(x + 1) + 3(x + 1)
 
   =   (x2 + 3)(x + 1)
  b)  2x3 − 6x2 + 5x − 15   =   2x2(x − 3) + 5(x − 3)
 
   =   (2x2 + 5)(x − 3)
  c)   3x3 − 15x2 − 2x + 10   =   3x2(x − 5) − 2(x − 5)
 
   =   (3x2 − 2)(x − 5)
  d)  12x3 + 2x2 − 18x − 3   =   2x2(6x + 1) − 3(6x + 1)
 
   =   (2x2 − 3)(6x + 1)
  e)  x3 + 2x2x − 2   =   x2(x + 2) − (x + 2)
 
   =   (x2 − 1)(x + 2)
  f)  12x3 − 6x2 − 2x + 1   =   6x2(2x − 1) − (2x − 1)
 
   =   (6x2 − 1)(2x − 1)

Problem 16.   Show by factoring the left-hand side:

(1 + x)2 + x(1 + x)2   =   (1 + x)3.
 
(1 + x)2 + x(1 + x)2   =   (1 + x) (1 + x)2
 
      (1 + x)2 is the common factor;
 
    =   (1 + x)3

Example 6.    Solve for x (Lesson 9):

pxq = rx + s
 
      1.  Transpose the x's to the left and everything else to the right:
pxrx = s + q
      2.  Factor:
x(pr) = s + q
      3.  Solve for x:
x = s + q
pr

Problem 17.   Solve for x.

ax + bx = c
 
x(a + b) = c
 
x =    c   
a + b

Problem 18.   Solve for x.

ax + b = cx + d
 
axcx = db
 
x(ac) = db
 
x = db
ac

Problem 19.   Solve for x.

axa = x
 
axx = a
 
x(a − 1) = a
 
x =    a   
a − 1

Problem 20.   Solve for x.

2 log 3· x + 5 log 3 = x + log 2.

The student might think that this is a difficult problem. But it is the same as Problem 18 It has the same form as Problem 18, which has the simple arguments a, b, c, d.

ax + b = cx + d.

In this equation,

a = 2 log 3,  b = 5 log 3,   c =1,   d = log 2.

The steps to solve this equation will be exactly the same as solving Problem 18.  And the solution will have the same form.

x  =  log 2 − 5 log 3
   2 log 3 − 1

This illustrates that algebra is purely formal. To solve an equation depends only on how things look. Any equation that has the same form as Problem 18 will have the same solution.  It is not necessary to know what "log 2" or "log 3" mean.

Back to Section 1

end

Next Lesson:  Multiplying binomials

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