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Factor Theorem

A container van has the function V (x) = 2x3-7x2 + 2x + 3, where x is the length of the container and (x-3) is a factor of the function. If, for instance, x=12, how do you calculate the container van’s dimensions?

 

This situation is an illustration of a real–world application that the factor theorem can solve.

Before we answer this, let us grasp the definition and procedure of the factor theorem. 

What is the Factor Theorem?

Factor Theorem

When is a polynomial with degree and is any real number then,

  • if, is a factor of; or
  • if is a factor of.

The fact that we get f (c)= 0 indicates that the remainder is zero and that ( x-c ) must be a factor of f (x) . Furthermore, ” c ” is a root of the polynomial if and only if f (c) = 0. 

Let us see how the factor theorem might apply to this.

Using factorization, if, for instance, we have the polynomial  x2 + 3x – 28, we have ( x  – 4) and ( x +7 ) as factors its factors. We can say that in this case c = 4 from the factor (x-4) while c = -7 from the factor ( x+7 ).

We must obtain f (c) =0 when substituting the values c = 4 and c =- 7 to f (x) to show that ( x-4 ) and ( x+7) are factors of the polynomial f (x) = x2 + 3x – 28 using the factor theorem.

Since f (4) = 0 and f (- 7) = 0, then (x-4) and (x+7) are both factors of the polynomial f x = x2 + 3x – 28.

Proof of Factor Theorem

Let us review that if f x is a polynomial and ( x-c  ) is its factor, then f (c) = 0. Therefore, when a polynomial is divided by ( x-c ), we will get the quotient h ( x ) and the remainder r.

Consequently, by division algorithm we can write f ( x ) = h (x)( x-c )+r 

According to the remainder theorem, the remainder of a polynomial having the root x=c is equal to r, that is,  f (c) = r.

Therefore, we could have:

If (x-c) is a factor of f (x) , then the remainder must be 0 upon division of f (x) by  (x-c) . That is,

f (x) = h (x)( x-c )+f (c) f (x) = h (x)( x-c )+0 f (x) = h (x)( x-c)

We now have the following with x=c:

                                                          f (c) = h (c)( c-c )

                                                        f (c) =h (c)(0)

                                                        f (c) =0

Hence, it is clear from here that ( x-c ) is a factor of the polynomial f (x) .

Synthetic Division and Factor Theorem

We shall obtain f ( c )= 0 if f ( x ) is a polynomial function and ( x-c ) is a linear factor of f ( x ). This implies that if we divide f ( x ) by ( x-c ), then the remainder is zero when f ( c )=0. As such, let us consider using synthetic division as an alternative method to determine whether a linear factor ( x-c ) is a factor of f ( x ).

Let us say for example we want to show using synthetic division that ( x-5 ) is a factor of the polynomial x4-6x3-9x2+ 94x-120.

Solution:

Step 1: We will use c = 5 from the given divisor ( x-5 ). The coefficients of the dividend x4– 6x3– 9x2+ 94x- 120  should be listed across the top.

Step 2: Bring down 1 since it is the leading coefficient.

Step 3:  Multiply 1 by five and write the product 5 in the middle row.

Step 4:  Add – 6 and 5 and write the sum in the bottom row.

Step 5:  Repeat the process of multiplying the number in the bottom row to the value of c=5 and adding the product to the number in the first row. 

The last number at the bottom, which is zero, gives the remainder. Since the remainder is zero, ( x-5 ) is a factor of x4– 6x3– 9x2 + 94x -120.

Examples

Example 1

Show that ( x-4 ) is a factor of x3-x2-22x +40.

Solution:

Using the factor theorem, let us use c = 4.  

Now substitute 4 to the given polynomial x3-x2-22x +40.

f ( 4 ) = ( 4 )3 – ( 4 )2 – 22 ( 4 ) + 40f ( 4 ) = 64 – 16 – 88 + 40f ( 4 ) = 0

Since f ( 4 ) = 0, then ( x – 4 ) is a factor of  x3-x2-22x +40.

Example 2

Check whether ( x+6 ) is a factor of x3+ 4x2 – 20x – 48.

Solution:

Using c= -6,  we have

f ( – 6 ) = (- 6)3 +  4 (- 6 )2 -20 (- 6) – 48f (- 6)
            = – 216  + 4 (36) + 120 -48f (- 6)
            = – 216  + 144 +120 – 48f (- 6)
            = 0

Since f -6 = 0, then ( x+6 ) is a factor of  x3 + 4x2 – 20x -48. 

Example 3

Show that ( x-6 ) is a factor of x4 – 9x3 + 8x2+ 84x -144 using the factor theorem.

Solution:

Using the factor theorem, let us use c= 6.  

Now substitute 6 to the given polynomial x4– 9x3+8 x2 + 84x-144 to get f (6) .

f (6) = ( 6 )4 -9( 6 )3 + 8( 6 )2 + 84( 6 ) -144f ( 6 )
       = 1296 -9( 216 ) + 8( 36 ) + 504 -144f ( 6 )
        =1296-1944 +288 + 504 -144f ( 6 )
        = 0

Since f (6) =0, then ( x-6 ) is a factor of  x4– 9x3+8 x2 + 84x-144.

Example 4

Determine whether ( 2x+4 ) is a factor of  2x3 + 12x2 +22x +12.

Solution:

Let us first determine what value of x  would make 2x+4 equal to 0.

Hence,

$2x+4\: =\: 02x\: =\: -4x\: =\: \frac{-4}{2}x\: =\: -2$

To determine whether 2x+4 is a factor of 2x3 + 12x2 + 22x + 12, let us try to substitute c=- 2. Thus,

f (- 2 ) = 2( – 2 )3+ 12( – 2 )2 + 22( – 2 ) +12f ( – 2 ) = -16 + 48 – 44 + 12f ( – 2 ) =  0

Since f (- 2 ) = 0, ( 2x+4 ) is a factor of  2x3 + 12x2 + 22x + 12.

Example 5

Show that ( 3x-1 ) is a factor of 3x4-25x3+ 59x2– 47x +10.

Solution

Let us first find out the value of x at which 3x-1 equals 0. 3x = 1x = $\frac{1}{3}$

To check whether 3x-1 is a factor of 3x4-25x3+59x2-47x +10, let us try to use the value c = $\frac{1}{3}$.

$f(\frac{1}{3})=3(\frac{1}{3})^4 – 25(\frac{1}{3})^3 + 59(\frac{1}{3})^2 – 47(\frac{1}{3}) +10f (\frac{1}{3})$
                      = $3(\frac{1}{81}) -25 (\frac{1}{27}) + 59( \frac{1}{9} ) – \frac{47}{3} + 10f (\frac{1}{3})$
                      = $\frac{3}{81} – \frac{25}{27} + \frac{59}{9} – \frac{47}{3} + 10f (\frac{1}{3})$
                      =  0

Since f ($\frac{1}{3}$)  =0, then ( 3x-1 ) is a factor 3x4-25x3+59x2-47x +10.

Example 6

Determine whether ( x-3 ) is a factor of x3– 10x2+ 11x+ 70.

Solution:

Let us solve for f (c) using the factor theorem. c = 3 based on the given.

So let us use c=3 to determine whether ( x-3 ) is a factor of x3– 10x2+ 11x+ 70.

f ( 3 ) = ( 3 )3 -10( 3 )2 + 11( 3 ) + 70f( 3 ) = 27 – 10( 9 ) + 33 + 70f( 3 ) = 27- 90+ 33+ 70f( 3 )= 40

Since f (3) ≠0 , then ( x-3 ) is not a factor of  x3– 10x2+ 11x+ 70.

Example 7

Show that ( 2x+1 ) is a factor of 2x4+7x3-9x2-22x-8.

Solution:

Let us first find out the value of x at which 2x+1 equals 0. 2x =  -1     x = – 1/2

To check whether 2x+1 is a factor of 2x4+7x3-9x2-22x-8,  let us try to use the value c =- 1/2.

f (-1/2 ) = 2 (- 1/2 )4 +7( -1/2 )3 -9(-1/2)2 – 22(-1/2) -8f (-1/2)  = 2 (1/16) +7 (-1/8) – 9 (1/4) + 22/2 – 8f – (1/2) = 2/16 – 7/8 – 9/4  + 11 -8f – 1/2 =  0

Since f – 12  =0, then ( 2x +1  ) is a factor 2x4+7x3-9x2-22x-8.

Example 8

Using the factor theorem, find the remainder when the polynomial x3– 20x2+ 123x- 216  is divided by (x-2) .

Solution:

To find the remainder using the factor theorem, we will solve for the value of f (c) . If we let f (x) = x3– 20x2+ 123x- 216 and c = 2 then,

f (2) = ( 2 )3-20( 2 )2 +123( 2 )-216f( 2 )= 8 – 20( 4 ) + 246 – 216f( 2 ) = 8 – 80 + 246 -216f( 2 ) = – 42

Therefore, – 42 is the remainder when x3– 20x2+ 123x- 216 is divided by  ( x-2 ).

Example 9

Determine if the polynomial f (x) = x3-6x2-51x+280 is divisible by (x-8) .

Solution:

Let us solve for f (c) using the factor theorem. From the given divisor, we have c = 8 

To show that f (x) = x3-6x2-51x+280 is divisible by ( x-8 ), we must obtain f (8) = 0.

Since f (8) =0, then f (x) = x3-6x2-51x+280 is divisible by ( x-8 ).

Example 10

Show that x3+ 2x2-11x-12 is divisible by x2+ 5x+ 4 using factor theorem.

Solution:

The factors of the divisor x2+ 5x +1 are ( x + 1 ) and ( x + 4 ).

Thus, using the factor theorem to show that x3+ 2x2– 11x- 12 is divisible by x2+5x+ 4, we must obtain f (-1) = 0 and f (-4) =0. 

Since f (-1) =0 and f (-4) =0, then x3+ 2x2– 11x- 12 is divisible by x2+ 5x+ 4.

Example 11

If  ( x+6 ) is a factor of f (x) = x3+ 13x2+ kx+ 60, find the value of k.

Solution:

Considering that (x+6) is a factor of the given polynomial, we must find the value of x that will make it equal to zero. Thus, 

              x + 6 =0

              x = -6

To solve for the value of k, let us substitute x =-6 to the polynomial f (x) = x3+ 13x2+kx+ 60. 

(-6 )3 + 13( -6 )2+ k( -6 ) + 60
= 0 – 216 + 13( 36 ) + k(-6 )+ 60 
= 0 – 216 + 468 -6k +60
= 0- 6k +312 
0-6k  =  -312k
=  -312-6k
K = 52

Therefore, for ( x+6 ) to be a factor of the polynomial x3+ 13x2+kx+ 60, k must have a value of 52. 

Example 12

Show using synthetic division that (x+3) is a factor of f x =x3– 6x2 -13x + 42.

Solution:

Since the given linear divisor is ( x+ 3), we will use c =-3. In the top row of the synthetic division setting, the dividend coefficients will be written. 

Since the final number in the bottom row of the synthetic division process, the remainder, is zero, this shows that ( x+3 ) is a factor of f x =x3– 6x2 -13x + 42.

Example 13

Using synthetic division, show whether (x-3) is a factor of f (x) =x3– 12x2+44x-48.

Solution:

We will use c=3 since the given linear divisor is ( x-3 ). In the top row of the synthetic division setting, the dividend coefficients will be written. 

The final number in the bottom row of the synthetic division process shows that the remainder is 3. Therefore, ( x-3 ) is not a factor of f x =x3– 12x2+44x-48.

Example 14

Which of the following polynomial functions has the linear factor ( x-5 ) ?

f (x) = x3– 6x2 – 37x+210p (x) = x3– 5x2 – 41x+ 45j (x) = x3– 4x2 – 77x  + 360

Solution:

To check whether each function has the linear factor (x-4) , we will use x =4 and solve for the values of f (x) ,  p (x) , and j (x) . 

The linear factor ( x-5 ) is a factor of  f (x) =x3– 15x2 + 59x – 45 and j x =x3– 4x2 – 77x  + 360  since both of these functions are equal to zero when x= 5. On the other hand, since p (5) is not equal to zero,  p (x) = x3– 5x2 – 41x + 45 does not include (x-5) as its factor 

Let us now try to answer the real-life problem raised earlier in the introduction to this topic since we have already established the concept and methods of the factor theorem. The next illustration shows how to apply the factor theorem to get the dimensions of the given figure.

Example 15

A container van has the function V (x) =2x3-7x2 + 2x +  3, where x is the length of the container and (x-3) is a factor of the function. If, for instance, x = 12, how do you calculate the container van’s dimensions?

Solution:

By applying the factor theorem, to prove that ( x-3 ) is a factor of the function V (x) =2x3– 7x2 + 2x + 3, we must determine the value of V ( 3 ) by substituting x = 3 to the function. Thus,

V (3) = 2( 3 )3– 7( 3 )2 + 2( 3 )+3V( 3 ) = 2( 27 ) -7( 9 ) + 6 + 3V( 3 ) = 54- 63 + 6 + 3 V( 3 ) = 0. 

Given that V (3) = 0, ( x-3 ) is a factor of V (x) =2x3– 7x2 + 2x + 3.

Now let us use synthetic division to determine the other factors of the function aside from ( x-3 ). The diagram below shows the results of dividing V (x) = 2x3– 7x2+ 2x+3 by the factor ( x-3 ). The value x= 3 comes from the divisor ( x-3 ). The coefficients of the dividend V (x) = 2x3-7x2+2x + 3 gave the integers 2, -7,  2, and 3, respectively.

The numbers 2 , -1 , and -1 give the coefficients of the quotient, that is , ( 2x2-x-1 ). Therefore, V (x) =2x3-7x2+ 2x + 3 has the factors (x-3) and (2x2-x-1) . 

In equation, 

V (x) = (x-3 )( 2x2-x-1 )

Using the binomial factors of ( 2x2-x-1 ), we have:

V (x) = ( x-3 )( 2x+1 )( x-1 )

  To determine the dimensions of the container van, substitute the given value x=12 with the factors of V (x) =2x3-7x2+ 2x +3.

Hence, the dimensions of the container van are 9 ft. x  25 ft. x  11 ft.

Summary

  • According to the factor theorem, when c is any real number and f(x) is a polynomial with degree n≥1, then, (x-c) is a factor of f (x) if f (c)=0.
  • (c)=0 indicates that the remainder is zero and that (x-c) must be a factor of f (x)
  • ” c ” is a root of the polynomial if and only if f( c )=0.
  • Let us say that we want to show ( x-5 ), ( x-2 ), and ( x-6 ) are all factors of the polynomial function f(x)=x3– 13x2+ 52x-60. Using factor theorem, we need to use c=5, c=2, and c=6. Thus,

    when c =5
    f (5) = (5 )3 -13( 5 )2+52( 5 )-60f( 5 ) =125 -13( 25 )+ 260 – 60f( 5 )=125 – 325 + 260 – 60f( 5 )=0

    when c =2
    f (2) = ( 2 )3– 13 ( 2 )2 +52 (2) -60f( 2 )= 8 – 13( 4 ) + 104 – 60f( 2 )= 8 -52 + 104 – 60f( 2 )=0

    when c =6
    f (6) =( 6 )3-13 ( 6 )2 + 52( 6 ) -60f( 6 ) = 216 -13( 36 ) + 312 – 60f( 6 ) = 216 – 468 + 312- 60f( 6 )=0

    Since we have proven that f(c) =0 when we plug in the values of c which are 2 , 5 and 6 to the given polynomial, then (x-5), ( x-2 ), and ( x-6 ) are all factors of f (x) = x3 – 13x2 + 52x – 60.  
  • To demonstrate that a linear binomial is a factor of a particular polynomial function, another approach is to use synthetic division.
  • Both the remainder theorem and the factor theorem are helpful techniques, without the use of long division, for figuring out the remainder of a given polynomial and determining whether a linear function is a factor.

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