Question

Find factors of $ {z^3} - 10{z^2} + 58z - 136 = 0 $

Answer 1

Hint: The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. In a polynomial equation, the highest exponent of the polynomial is called its degree. And according to the Fundamental Theorem of Algebra, a polynomial equation has exactly as many roots as its degree. The roots of an equation can be found out by factoring the equation and also by a special formula called the quadratic formula. In the given question, we have to find one root by hit and trial method and the other roots by the quadratic formula.

Complete step-by-step answer:
Now, the above equation has degree three, so it has three solutions. The equation can’t be solved by factorization easily, so we use the hit and trial method for finding one of the roots of the equation.
At z=1,
 $
  f(x) = {(1)^3} - 10{(1)^2} + 58(1) - 136 \\
  f(x) = 1 - 10 + 58 - 136 \\
  f(x) = - 87 \\
   \Rightarrow f(x) \ne 0 \\
  $
So, \[z - 1 = 0\] is not a factor of the given equation.
At z=2,
 $
  f(x) = {(2)^3} - 10{(2)^2} + 58(2) - 136 \\
  f(x) = 8 - 40 + 116 - 136 \\
  f(x) = - 52 \\
   \Rightarrow f(x) \ne 52 \;
  $
So, \[z - 2 = 0\] is not a factor of the given equation.
At z=3,
 $
  f(x) = {(3)^3} - 10{(3)^2} + 58(3) - 136 \\
  f(x) = 27 - 90 + 174 - 136 \\
  f(x) = - 25 \\
   \Rightarrow f(x) \ne 0 \;
  $
So, $ z - 3 = 0 $ is not a factor of the given equation.
At z=4,
 $
  f(x) = {(4)^3} - 10{(4)^2} + 58(4) - 136 \\
  f(x) = 64 - 160 + 232 - 136 \\
   \Rightarrow f(x) = 0 \;
  $
So, $ z - 4 = 0 $ is a solution of the given equation. Using this we can write the given equation $ {z^3} - 10{z^2} + 58z - 136 = 0 $ as –
 $
  {z^3} - 4{z^2} - 6{z^2} + 24z + 34z - 136 = 0 \\
   \Rightarrow {z^2}(z - 4) - 6z(z - 4) + 34(z - 4) = 0 \\
   \Rightarrow (z - 4)({z^2} - 6z + 34) = 0 \\
   \Rightarrow {z^3} - 10{z^2} + 58z - 136 = (z - 4)({z^2} - 6z + 34) \\
   \Rightarrow f(x) = (z - 4)({z^2} - 6z + 34) \;
  $
So, to find the factors, we get –
 $ {z^2} - 6z + 34 = 0 $
Solving the above equation by completing the square method –
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  z = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4 \times 1 \times 34} }}{{2(1)}} \\
  z = \dfrac{{6 \pm \sqrt { - 100} }}{2} \\
  z = \dfrac{{6 \pm 10i}}{2} \\
   \Rightarrow z = 3 \pm 5i \;
  $
So, $ z - 3 - 5i = 0 $ and $ z - 3 + 5i = 0 $ are the other two factors.
Hence, the factors of the equation $ {z^3} - 10{z^2} + 58z - 136 = 0 $ are $ (z - 4) $ , $ z - 3 - 5i = 0 $ and $ z - 3 + 5i = 0 $ .
So, the correct answer is “ $ (z - 4) $ , $ z - 3 - 5i = 0 $ and $ z - 3 + 5i = 0 $ ”.

Note: While finding the roots by hit and trial method, the answer was decreasing towards zero as we were increasing the value of z. If the answer doesn’t come close to zero after two to three trials then we can skip to bigger values of z to get closer to zero. $ \sqrt { - 1} $ is an imaginary number and is called iota, denoted by $ i $ , thus the two roots of the given equation are imaginary.