Question

How do you find the zeros of $ P(x) = {x^4} - 2{x^3} - 7{x^2} + 8x + 12 $ ?

Answer 1

Hint: A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0 (or make Y=0). It is an X-intercept. The root is the X-value, and zero is the Y-value. It is not saying that imaginary roots = 0.

Complete step by step solution:
We will use the factor theorem for solving the above biquadratic equation, the equation is called to be as the biquadratic when the degree of the equation is $ 4 $ . According to the factor theorem we write,
For $ f\left( x \right) $ a polynomial which is easily divisible by \[\left( {x - a} \right)\], then we can write
 $ f\left( a \right) = 0 $ ,and the value of $ a $ becomes the root of the polynomial $ f\left( x \right) $ .
We can write the $ P\left( x \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)\left( {x - {a_3}} \right)\left( {x - {a_4}} \right) $
Also the values of $ {a_1}{a_2}{a_3}{a_4} $ are factors of the constant term here which is $ 12 $ that means it can be among
 $ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 $ . These values are the factors of the constant term in this equation.
We first put $ a = 1 $
 $ P\left( x \right) = 1 - 2 - 7 + 8 + 12 = 12 $
Is not a factor because it is not zero.
 We now put $ a = - 1 $
 $ P\left( x \right) = 1 + 2 - 7 - 8 + 12 = 0 $
Thus $ - 1 $ is a factor.
When putting $ a = 2 $ we get,
 $
  P\left( x \right) = 1 \cdot {2^4} - 2 \cdot {2^3} - 7 \cdot {2^2} + 8 \cdot 2 + 12 \\
  P(x) = 16 - 16 - 28 + 16 + 12 = 0 \;
\ $
Thus $ a = 2 $ is also a factor
On putting $ a = - 2 $
We can write the equation as,
 $
  P\left( x \right) = 1 \cdot {\left( { - 2} \right)^4} - 2 \cdot {\left( { - 2} \right)^3} - 7 \cdot {\left( { - 2} \right)^2} + 8 \cdot \left( { - 2} \right) + 12 \\
  P(x) = 16 + 16 - 28 - 16 + 12 = 0 \;
  $
Thus we get another factor in the form of $ a = - 2 $
And last factor will be,
 $ \dfrac{{12}}{{ - 1 \cdot 2 \cdot - 2}} = 3 $
This can be checked by putting the value of the root above into the given equation,
 $
  P\left( x \right) = 1 \cdot {3^4} - 2 \cdot {3^3} - 7 \cdot {3^2} + 8 \cdot 3 + 12 \\
  P(x) = 81 - 54 - 63 + 24 + 12 = 0 \;
 $
Thus the above value is verified .
Thus we have found all the four roots of the equation which are ,
 $ - 2, - 1,2,3 $ .
It is also worth noting that all the factors here are the factor of the constant term in the equation.

Note: Always remember the factor theorem we will use this theorem in other questions as well so it is very important theorem in the context :
For $ f\left( x \right) $ a polynomial which is easily divisible by \[\left( {x - a} \right)\], then we can write
 $ f\left( a \right) = 0 $ ,and the value of $ a $ becomes the root of the polynomial $ f\left( x \right) $ .