The given equation is:
x^4-8x^3+5x^2+14x
In this question, we have to find the zeros algebraically. Hence, we need to find the values of x that satisfy the equation.
Note: The roots (values of x) are equal to the power of the equation. Therefore, the given equation will have 4 roots
Therefore,
x^4-8x^3+5x^2+14x = 0
or
x(x^3-8x^2+5x+14) = 0
=> x = 0 and x^3-8x^2+5x+14 = 0
Therefore, the first root is x = 0.
Now,
x^3-8x^2+5x+14 = 0
To simplify that, try putting different values of x (starting from 1 and -1). The value that makes both sides equal to zero would be the solution.
Let's start with (1):
For x = 1
(1)^3 - 8(1)^2 + 5(1) + 14 = 0
1 -8(1) + 5 + 14 = 0
-1 -8 +5 + 14 = 0
- 9 + 19 = 0
10 = 0
It is false, therefore, x = 1 is not the solution.
For x = -1
(-1)^3 - 8(-1)^2 + 5(-1) + 14 = 0
-1 -8(1) - 5 + 14 = 0
-1 -8 -5 + 14 = 0
- 14 + 14 = 0
0 = 0
hence, x = -1 is the second solution.
If x = -1 is the solution, (x + 1) should be the factor of x^3-8x^2+5x+14. Therefore, the whole equation can be divided by factor
x^3-8x^2+5x+14 / (x+1). = x^2 - 9x + 14. (using division)
Now, let's find the roots of x^2 - 9x + 14
x^2 - 9x + 14 = 0
x^2 - 7x - 2x + 14 = 0
x(x - 7) - 2 (x - 7) = 0
(x - 7) (x - 2) = 0
or
x - 7 = 0. and x - 2 = 0
x = 7 and x = 2
Therefore, the zeros of the given equation are, 0, -1, 2 and 7.