Factorising Polynomials

I ask Johnny to find the factors of the cubic polynomial, f(x) = 1x³ – 2x² – 7x – 4.

He begins by noting that the highest power of x is 3 or equivalently f(x) is of degree ‘3’ and consequently f(x) must have three factors.  He decides to use the factor theorem: If f(α) = 0, then (x – α) is a factor of f(x).

Given that the constant term of f(x) is -4, he chooses ±1, ±2, or ±4 as candidates for α.

Is (x – 1) a factor?  f(1)  = 1(1)³ – 2(1)² – 7(1) – 4 = 1 – 2 – 7 – 4 = -13                  

Is (x + 1) a factor? YES.  f(-1)  = 1(-1)³ – 2(-1)² – 7(-1) – 4 = -1 – 2 + 7 – 4 = 0               

Is (x – 2) a factor?  f(2)  = 1(2)³ – 2(2)² – 7(2) – 4   = ⁺8 – large negative ≠ 0         

Is (x + 2) a factor?  f(-2)  = 1(-2)³ – 2(-2)² – 7(-2) – 4 = -8 – 8 + 14  – 4 ≠ 0            

Is (x – 4) a factor? YES.  f(4)  = 1(4)³ – 2(4)² – 7(4) – 4 = 64 – 32 – 28 – 4 = 0              

Is (x + 4) a factor?  f(-4)  = 1(-4)³ – 2(-4)² – 7(-4) – 4 = -64 – 32 +2 8 – 4 ≠ 0              

Having used all of his candidates for α, Johnny has found only two factors for f(x).  Because he knows there must be a third factor, he decides to apply polynomial identities.

f(x) = (x +1) (x – 4) (bx + c) = 1x³ – 2x² – 7x – 4.  By inspection of the cubic term, b must equal 1.  Inspection of the constant term reveals that c = 1 also.

Johnny’s solution is  f(x) = 1x³ – 2x² – 7x – 4 = (x +1) (x – 4) (x + 1) =  (x +1)² (x – 4). 

Mary, in the same class, dislikes ‘guesswork,’ and avoids the factor theorem as much as possible.  Like Johnny, she finds (x + 1) using the factor theorem, but thereafter she continues with polynomial long division.

 

As an aside, if you ever want to check that your long division homework is correct, WolframAlpha can be of assistance: www.wolframalpha.com.

Because the quotient is 1x² – 3x –  4, she knows that f(x) = (x+ 1) (1x² – 3x –  4).  She factorises the quadratic term using mental maths: x² – 3x –  4 = (x – 4) (x + 1).  She and Johnny are in agreement, f(x) = (x +1) (x – 4) (x + 1) =  (x +1)² (x – 4).

Mr. Patterson gives both students gold stars and assigns a new problem. Solve f(x) = g(x), with f(x) as before and g(x) = x² – 3x –  4.

Both students agree that f(x) = g(x) ought to be re-written as f(x) – g(x) = 0. 

Johnny proposes to work with both polynomials in expanded form and to simplify as follows:

f(x)  – g(x) = (1x³ – 2x² – 7x – 4) – (x² – 3x –  4)

= 1x³ – 3x² – 4x

= (x) (x² – 3x –  4)

= (x) (x – 4) (x + 1).   

Mary has a different method.  She will work with both polynomials in factored form.  Since she recognizes that g(x) is the quotient she calculated earlier, she knows g(x) has common factors with f(x).  To make the factorization clearer, she underlines the common factors and rewrites g(x) as g(x) × 1.

Thus f(x)  – g(x)

= f(x)  – g(x) × 1

= (x +1) (x – 4) (x + 1) – (x +1) (x – 4) (1)

= [(x +1) (x – 4)] [(x + 1) – 1]

 = (x +1) (x – 4) (x)

Mary and Johnny both know the zero law, “When the product is zero, at least one of the factors must be zero.”  Symbolically, a × b = 0 requires either a = 0 or b = 0.

Consequently they write:

0= (x +1) (x – 4) (x)

0 = x + 1  so  x = -1    OR      0 = x – 4  so  x = 4      OR      0 = x 

Again both students are correct.

For you to solve:

y =  f(x)           ①

y = 6x – 6        ②

f(x) is still the same cubic polynomial, 1x³ – 2x² – 7x – 4.

h(x) = 6x – 6 is a binomial of degree one, i.e., it represents a line.

Summing up:  

Mary and Johnny’s problem is from Exercise 4.6 of Panpac Additional Mathematics by Ho Soo Theng and Khor Nyak Hiong.

Mary and Johnny are not actual students in my class; they are pedagogical prototypes.  For this particular problem, although both students had correct answers, Mary demonstrated competence in a greater variety of mathematical methods.