Question

Find all zeroes of the polynomial \[2{x^4} - 9{x^3} + 5{x^2} + 3x - 1\] if two of its zeroes are \[2 + \sqrt 3 \]​ and \[2 - \sqrt 3 \]​.

Answer 1

Hint: If you're dividing a polynomial by something more complicated than just a simple monomial (that is, by something more complicated than a one-term polynomial), then you'll need to use a different method for the simplification. That method is called "long polynomial division", and it works just like the long (numerical) division you did back in elementary school, except that now you're dividing with variables. We will also be using factoring and simplifying the algebraic expressions to help us arrive at different zero roots of the equation.

Complete step-by-step answer:
Given in the question: -
\[f(x) = 2{x^4} - 9{x^3} + 5{x^2} + 3x - 1\]
Zeroes = \[2 + \sqrt 3 \]​ and \[2 - \sqrt 3 \]​
Since, the equation is a polynomial of a 4th order so there will be four zeroes, say, a, b, c and d.
The Factor Theorem
If (x – a) is a factor of the polynomial f(x), then f(a) = 0
In other words, if a polynomial f(x) can be divided by (x - a) without a remainder, then x = a is a root of f(x) (so f(a) = 0).
The factor theorem is important because it can be an easy way of finding factors that would otherwise be difficult to find.
Given the zeros, we can write the factors = \[\left( {x{\text{ }}-{\text{ }}2{\text{ }} + {\text{ }}\sqrt 3 } \right)\]and \[\left( {x{\text{ }}-{\text{ }}2{\text{ - }}\sqrt 3 } \right)\]
{Since, if x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}
Multiplying these two factors, we can get another factor which is:
\[
  \left( {x{\text{ }}-{\text{ }}2{\text{ }} + {\text{ }}\sqrt 3 } \right)\left( {x{\text{ }}-{\text{ }}2{\text{ - }}\sqrt 3 } \right) \\
   = {\left( {x - 2} \right)^2} - 3 \\
   = {x^2} - 4x + 4 - 3 \\
   = {x^2} - 4x + 1 \\
 \]
So, dividing f(x) with \[{x^2} - 4x + 1\]
\[
  {x^2} - 4x + 1\mathop{\left){\vphantom{1{2{x^4} - 9{x^3} + 5{x^2} + 3x - 1}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^4} - 9{x^3} + 5{x^2} + 3x - 1}}}}
\limits^{\displaystyle \,\,\, {2{x^2} - x - 1\,\;\quad \quad \quad }} \\
  \quad \quad \quad \quad \;\,2{x^4} - 8{x^3} + 2{x^2} \\
  \quad \quad \quad \quad \;\,\dfrac{{ - \;\quad + \quad \quad - \quad \quad }}{
  \quad \quad - {x^3} + 3{x^{^2}} + 3x \\
  \quad \quad - {x^3} + 3{x^{^2}} + 3x \\
  \quad \quad \;\dfrac{{ + \quad - \quad - \quad \quad \quad \quad }}{
  \;\;\quad \; - {x^2} + 4x - 1 \\
  \;\;\quad \; - {x^2} + 4x - 1 \\
  \dfrac{{\quad \quad \quad + \quad - \quad + \quad }}{0} \\
 }\quad \\
 } \\
 \]
\[f\left( x \right){\text{ }} = {\text{ }}\left( {2{x^2} - x - 1} \right){\text{ }}\left( {{x^2} - 4x + 1} \right)\]
Now, solving\[\left( {2{x^2} - x - 1} \right)\], we get the two remaining roots as
Factoring the above equation, we get,
\[ = \left( {2{x^2} - 2x + x - 1} \right)\]
\[ = \left( {2x(x - 1) + (x - 1)} \right)\]
\[ = \left( {2x - 1)(x - 1} \right)\]
\[ \Rightarrow x = \dfrac{1}{2},1\]

Zeros of the polynomial \[ \Rightarrow x = \dfrac{1}{2},1,2 + \sqrt 3 ,2 - \sqrt 3 \]

Note: If the divisor is a factor of the dividend, you will obtain a remainder equal to zero. If the divisor is not a factor of the dividend, you will obtain a remainder whose index is lower than the index of the divisor. With algebraic long division, practise makes perfect- the best way to learn how to do them properly is to do loads of examples until you get them right every time.