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Common factor:  2nd Level

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The leading term

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,

the leading term in 3x².

We like the leading term to be positive.  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 + 2 + xn + 3 = xn(x² + x3)   b)   xn + 2 + xn + 4 = xn(x² + 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 + 2 = 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 + 2 + xn + 3 = xn + 2(1 + x)
 
   c)   xnxn − 2 = xn − 2(x² − 1)   d)   xn − 1xn + 1 = xn − 1(1 − x2)
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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 13.   See the common factor.

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 14.   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 + 2x² − x − 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 15.   Show by factoring the left-hand side:

(1 + x)² + x(1 + x   =   (1 + x)3.
 
(1 + x)² + x(1 + x   =   (1 + x) (1 + x
 
      (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 16.   Solve for x.

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

Problem 17.   Solve for x.

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

Problem 18.   Solve for x.

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

Back to Section 1


Next Lesson:  Quadratic Trinomials


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