Question

How do you find the number of roots for $ F(x) = {x^3} - 10{x^2} + 27x - 12 $ using the fundamental theorem of algebra?

Answer 1

Hint: The fundamental theorem of algebra suggests that any non - constant polynomial with complex or possible real coefficients has zero in the complex numbers. Zeroes are also known as the roots of the polynomial. Here we will find the zeroes of the polynomial by the theorem and calculations.

Complete step by step solution:
By theorem, if the polynomial of the degree is $ n > 0 $ then the polynomial will have zeros in the complex numbers counting multiplicity.
Degree is defined as the highest power of the polynomial.
Here, the degree of the given polynomial is $ 3 $ . So, it has three roots. It may have all the three positive real roots or the combinations of the real and a pair of non-real complex conjugate roots.
Now, by using the rational root theorem rational roots must be expressed in the form of $ \dfrac{p}{q} $ .
So, here for the integers p, q with p a divisor with the constant term $ ( - 12) $ and q a divisor of the coefficient $ 1 $ of the first term.
The only possible roots are: $ 1,2,3,4,6,12 $
Now, by using trial and error method we get
 $ f(4) = 0 $
And therefore $ (x - 4) $ is a factor.
So, the given expression can be re-written as –
  $ {x^3} - 10{x^2} + 27x - 12 = (x - 4)({x^2} - 6x + 3) $ (By division method)
Now, to get the factors of the term $ ({x^2} - 6x + 3) $ use standard quadratic equation –
Compare the above equation with the standard quadratic equations- $ a{x^2} + bx + c = 0 $ where roots will be defined as \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Also, $ \Delta = {b^2} - 4ac $
Place the values from the given comparison
 $ \Rightarrow \Delta = {( - 6)^2} - 4(1)(3) $
Simplify the above equation –
 $ \Rightarrow \Delta = 36 - 12 $
Do subtraction –
 $ \Rightarrow \Delta = 24 $
Take square root on both the sides of the equation –
 $ \Rightarrow \sqrt \Delta = \sqrt {4 \times 6} $
 $ \Rightarrow \sqrt \Delta = 2\sqrt 6 $
Now, the roots are \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Place the values in the above equation –
\[x = \dfrac{{ - 6 \pm 2\sqrt 6 }}{2}\]
Remove common factors from the numerator and the denominator.
\[x = - 3 \pm \sqrt 6 \]
Hence, the zeroes of the given polynomial are \[x = 4, - 3 \pm \sqrt 6 \]

Note: Be careful about the sign convention while simplification. Before using the standard quadratic formula, you can try once if you can simplify the given polynomial by using the split middle term method. Be good in factorization which is to get the terms which give the original number when multiplied for simplification.