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Equivalent Polynomials


Polynomial expressions are characterised by the degrees of their terms.

A polynomial may be written as a sum of terms

$a_0+a_1x+a_2x^2+a_3x^3+...+a_nx^n$a0+a1x+a2x2+a3x3+...+anxn .

This is a polynomial of degree $n$n

Two polynomials are the same for all values of the variable if and only if they have the same degree and all the coefficients in corresponding terms are the same.

Polynomials may also be given in factorised form


This, again, is a polynomial of degree $n$n because if it were to be expanded, it would look like the previous form, a sum of terms with different powers of $x$x, and the highest power would be $n$n.

If these two forms were representations of the same polynomial, The coefficients in the expansion of the second version would have to be the same as the coefficients in the first version. Thus, you can easily check that the coefficient of $x^n$xn in the second version is $k$k and so, $a_n=k$an=k.

Also, the constant term in the second version must be $r_1r_2r_3...r_n$r1r2r3...rn and this must be the same as $a_0$a0

Example 1

Given that the quadratic $2x^2+3x-5$2x2+3x5 is the same as the factorised form $(x-1)(ax+b)$(x1)(ax+b) can we discover $a$a and $b$b?

We expand the factorised form. Thus, $(x-1)(ax+b)=ax^2+(b-a)x-b$(x1)(ax+b)=ax2+(ba)xb. On comparing the quadratic and constant terms with those in the other given expression, we find that $a=2$a=2 and $b=5$b=5. Note that this is consistent with the coefficient of the linear term $b-a=3$ba=3.


In attempting to solve a polynomial equation, it can happen that we can see by inspection what one of the solutions is. For example, given $2x^3-5x^2+8x-5=0$2x35x2+8x5=0, we might notice that $x=1$x=1 is a solution.

If $x=1$x=1 is a solution, the factorised form of the polynomial must include the factor $x-1$x1. The other factor must be a quadratic. So, we can write


Now, if we could determine the coefficients $a$a, $b$b and $c$c we would be in a position to find the other roots of the cubic because we know how to find the roots of a quadratic. This is often done using a polynomial division algorithm but we can do it by equating coefficients as follows.

$(x-1)(ax^2+bx+c)$(x1)(ax2+bx+c) $=$= $ax^3+(b-a)x^2+(c-b)x-c$ax3+(ba)x2+(cb)xc


$a$a $=$= $2$2
$b-a$ba $=$= $-5$5
$c-b$cb $=$= $8$8
$c$c $=$= $5$5

This set of equations is consistent if $b=-3$b=3. The original equation can now be written $(x-1)(2x^2-3x+5)=0$(x1)(2x23x+5)=0

We look for solutions to $2x^2-3x+5=0$2x23x+5=0. In this case, there are no further real solutions because the discriminant is negative. This cubic has only one real root (and two complex roots). In other cases, we could proceed, using the quadratic formula or another method to find three real solutions to the cubic.


Example 2

A polynomial in a variable $x$x can be given an alternative form as a polynomial in $y=x+a$y=x+a where $a$a is a constant. Suppose we are given $f(x)=x^3+3x^2-x-1$f(x)=x3+3x2x1 and we wish to express this function in terms of the variable $y=x+5$y=x+5

The new function definition will look like $Ay^3+By^2+Cy+D$Ay3+By2+Cy+D. If the two polynomials are to be identically equal we have $A(x+5)^3+B(x+5)^2+C(x+5)+D\equiv x^3+3x^2-x-1$A(x+5)3+B(x+5)2+C(x+5)+Dx3+3x2x1 and we need to determine the coefficients $A$A, $B$B, $C$C and $D$D.

Since the polynomials are equal for all values of $x$x, they are equal for the particular value $x=-5$x=5. Substituting $x=-5$x=5 into the left-hand side causes all but the constant term to disappear and on the right we have $(-5)^3+3\times(-5)^2-(-5)-1=-46$(5)3+3×(5)2(5)1=46.

So, $D=-46$D=46.

The coefficient of the cubic term must be $A$A and this is $1$1 from the original form.

On expansion of the brackets, we see that the quadratic term is $3\times5Ax^2+B=3x^2$3×5Ax2+B=3x2. Since we have already determined $A$A, we deduce that $B=-12$B=12.

The constant term is $5^3A+5^2B+5C+D=-1$53A+52B+5C+D=1. So, $125-300+5C-46=-1$125300+5C46=1 and from this, we calculate that $C=44$C=44.

Thus, the equivalent form we seek is 

$f(x)=(x+5)^3-12(x+5)^2+44(x+5)-46$f(x)=(x+5)312(x+5)2+44(x+5)46 or



Example 3

Continuing the discussion above, we claim that a cubic in a variable $x$x can be expressed equivalently as a cubic in a variable $x+a$x+a and the constant $a$a can be chosen in such a way that the quadratic term in the new expression disappears. This is called a Tschirnhaus transformation. The idea was used in finding the general solution to the cubic polynomial equation by Tartaglia in the 16th century.

If we begin with $x^3+bx^2+cx+d$x3+bx2+cx+d, we might attempt to write this equivalently as $(x+a)^3+p(x+a)+q$(x+a)3+p(x+a)+q. How is this to be done?

In the second expression, there is a quadratic term when $(x+a)^3$(x+a)3 is expanded. Its coefficient is $3a$3a. So, $b=3a$b=3a and therefore $a=\frac{b}{3}$a=b3.

We now have $x^3+bx^2+cx+d\equiv\left(x+\frac{b}{3}\right)^3+p\left(x+\frac{b}{3}\right)+q$x3+bx2+cx+d(x+b3)3+p(x+b3)+q.


Here is a real example: $f(x)=x^3-9x^2+4x-1$f(x)=x39x2+4x1.

According to the above discussion, we can write this in the form $f(x)=\left(x-3\right)^3+p\left(x-3\right)+q$f(x)=(x3)3+p(x3)+q because $a=\frac{-9}{3}$a=93.

To determine $p$p and $q$q we expand this expression.

$f(x)$f(x) $=$= $\left(x-3\right)^3+p\left(x-3\right)+q$(x3)3+p(x3)+q
  $=$= $x^3-9x^2+27x-27+px-3p+q$x39x2+27x27+px3p+q
  $=$= $x^3-9x^2+x(27+p)-27-3p+q$x39x2+x(27+p)273p+q

Thus, $27+p=4$27+p=4 and $-27-3p+q=-1$273p+q=1. From this we deduce that $p=-23$p=23 and then $q=-43$q=43.

So, we have the equivalent expression,



Worked Examples

Question 1

Given that $ax^2-15x+25=\left(2x-5\right)\left(x-5\right)$ax215x+25=(2x5)(x5) for all values of $x$x, solve for $a$a.

Question 2

Consider the identity $x^3+9x^2+27x+29=A\left(x+3\right)^3+B$x3+9x2+27x+29=A(x+3)3+B for all real values of $x$x.

  1. Solve for $A$A.

  2. Solve for $B$B.

Question 3

$x^2+6x+11=A\left(x+3\right)^2+B$x2+6x+11=A(x+3)2+B for all real values of $x$x.

  1. Solve for $A$A.

  2. Solve for $B$B.




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