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Find the zeros algebraically: f(x)=x^4-8x^3+5x^2+14x

User Musannif Zahir
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1 Answer

12 votes
12 votes

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.

User LordAro
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