MHB Synthetic Division P(x)|2+3i= 0

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The discussion centers on using synthetic division to analyze the polynomial P(x) = x^4 - 4x^3 + 10x^2 + 12x - 39, given that 2 + 3i is a zero. It is noted that if a polynomial has a complex zero, its conjugate, 2 - 3i, is also a zero, leading to the conclusion that both are factors of P(x). The product of these factors results in a real quadratic polynomial, which can then be used for synthetic division. The process reveals that synthetic division is applicable for dividing by linear factors, while the quadratic factor can be divided without remainder. Ultimately, the discussion highlights the steps to find additional zeros of the polynomial through this method.
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P(x)= x^4-4x^3+10x^2+12x-39, using synthetic division given 2+3i is a zero of function
 
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shorty888 said:
P(x)= x^4-4x^3+10x^2+12x-39, using synthetic division given 2+3i is a zero of function

Exactly what do you want to do, what you have posted is not a question. Please post the question as asked.

CB
 
shorty888 said:
P(x)= x^4-4x^3+10x^2+12x-39, using synthetic division given 2+3i is a zero of function
If a polynomial with real coefficients has a complex zero, then the complex conjugate of that number is also a zero. Thus 2+3i and 2-3i are both zeros. By the factor theorem, $x-(2+3i)$ and $x-(2-3i)$ are both factors of $P(x)$. Hence so is their product $\bigl(x-(2+3i)\bigr)\bigl(x-(2-3i)\bigr)$. Work out that product (which is a real quadratic polynomial), then use synthetic division to divide $P(x)$ by that quadratic. The quotient will be another quadratic, which you can solve to get the other two zeros of $P(x)$.
 
My understanding of synthetic division is that it is used to divide by "x- a" for a constant a, not a quadratic. Of course, it is true that
(x-(2-3i))(x+(2- 3i))= ((x- 2)- 3i)((x-2)+ 3i)= (x-2)^2- (3i)^2= x^2- 4x+ 4+ 9= x^2- 4x+ 13 divides into x^4- 4x^3+ 10x^2+ 12x- 39 without remainder but synthetic division by x- (2+ 3i) is2+3i|1_____-4_______10_________12_______-39
__________2+3i_____-13_______-6+9i_______+39
____1____-2+3i______-3________6+9i________0

or x^3+ (-2+3i)x^2- 3x+ (6+ 9i)
 
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