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Question: How do you find all the zeroes of \[{x^4} - 4{x^3} + 14{x^2} - 4x + 13\] with the zero \(2 - 3i\)?...

How do you find all the zeroes of x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 with the zero 23i2 - 3i?

Explanation

Solution

According to the question we have to determine the zeros of the given expression x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 with the zero 23i2 - 3i. So, to determine the all the zeroes of the given polynomial expression x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 first of all we have to use the one of the zero which is 23i2 - 3i and as we all know that the all of the coefficients of the polynomial expression are real hence, we have to know that the complex conjugate of the zero will also be a zero. Therefore we have to determine another conjugate of the given zero which is 23i2 - 3i.
Now, since we have obtained both of the two zeroes so we have to obtain the factors of the polynomial expression which is x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13.
Now, we have to multiply both of the factors obtained to achieve the quadratic expression so that we can determine the other roots of the given polynomial expression.

Formula used: (a+b)(ab)=a2+b2................(A) \Rightarrow (a + b)(a - b) = {a^2} + {b^2}................(A)
(ab)2=a2+b22ab.............(B)\Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab.............(B)
i2=1.............(C)\Rightarrow {i^2} = - 1.............(C)
While solving the factors we have to apply the formula (C) as mentioned above to determine the quadratic expression.
Now, as we know that the quadratic expression is obtained with the help of the factors obtained so the quadratic expression will satisfy the given polynomial expression hence, we have to divide the given polynomial expression x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 by the quadratic expression obtained.

Complete step-by-step solution:
Step 1: First of all we have to use the one of the zero which is 23i2 - 3i and as we all know that the all of the coefficients of the polynomial expression are real hence, we have to know that the complex conjugate of the zero will also be a zero. Therefore we have to determine another conjugate of the given zero which is 23i2 - 3i.
\Rightarrow Conjugate of 23i2 - 3i is 2+3i2 + 3i
Step 2: Now, since we have obtained both of the two zeroes so we have to obtain the factors of the polynomial expression which is x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 which is as mentioned in the solution hint. Hence,
\Rightarrow (x(23i)(x - (2 - 3i) and (x(2+3i)(x - (2 + 3i)
Step 3: Now, we have to multiply both of the factors obtained to achieve the quadratic expression so that we can determine the other roots of the given polynomial expression. Hence,
(x(23i)×(x(2+3i) (x2+3i)×(x23i) \Rightarrow (x - (2 - 3i) \times (x - (2 + 3i) \\\ \Rightarrow (x - 2 + 3i) \times (x - 2 - 3i)
Step 4: Now, to solve the expression as obtained in the solution step 3 we have to use the formula (A) which is as mentioned in the solution hint. Hence,
(x2)2(3i)2\Rightarrow {(x - 2)^2} - {(3i)^2}
Step 5: Now, to solve the expression as obtained in the solution step 4 we have to use the formula (B) as mentioned in the solution hint. Hence,
x2+222×x×29i2 x2+44x9i2 \Rightarrow {x^2} + {2^2} - 2 \times x \times 2 - 9{i^2} \\\ \Rightarrow {x^2} + 4 - 4x - 9{i^2}
Step 6: Now, to solve the expression as obtained in the solution step 5 we have to use the formula (C) as mentioned in the solution hint. Hence,
x2+44x9(1) x2+44x+9 x24x+13 \Rightarrow {x^2} + 4 - 4x - 9( - 1) \\\ \Rightarrow {x^2} + 4 - 4x + 9 \\\ \Rightarrow {x^2} - 4x + 13
Step 7: Now, as we know that the quadratic expression is obtained with the help of the factors obtained so the quadratic expression will satisfy the given polynomial expression hence, we have to divide the given polynomial expression x44x3+14x24x+13{x^4} - 4{x^3} + 14{x^2} - 4x + 13 by the quadratic expression obtained. Hence,

x44x3+14x24x+13x24x+13 x2+1 \Rightarrow \dfrac{{{x^4} - 4{x^3} + 14{x^2} - 4x + 13}}{{{x^2} - 4x + 13}} \\\ \Rightarrow {x^2} + 1

Step 8: Hence, with the help of solution step 7 we can determine the remaining zeros of the polynomial expression with the factorx2+1{x^2} + 1. Hence,

x2+1=0 x2=1 \Rightarrow {x^2} + 1 = 0 \\\ \Rightarrow {x^2} = - 1

Now, with the help of the formula (A), as mentioned in the solution hint,
x=±i\Rightarrow x = \pm i

Hence, with the help of the formula (A), (B), and (C) we have determined the all of the zeroes of the given polynomial which are x=i,i,2+3i,23ix = i, - i,2 + 3i,2 - 3i.

Note: If one of the zero which is 23i2 - 3i and as we all know that the all of the coefficients of the polynomial expression are real hence, we have to know that the complex conjugate of the zero will also be a zero. The other root will be 2+3i2 + 3i
To determine the other zeroes it is necessary that we have to find the factors of the zeroes as obtained so that we can determine the other factors to find other zeros.