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Common factor:  2nd Level

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The leading term

Proof of the Rule of Signs

Factoring by grouping

Solving equations

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

Back to Section 1

end

Next Lesson:  Quadratic Trinomials

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