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- #1

- Thread starter shorty888
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- Thread starter
- #1

- Jan 26, 2012

- 890

Exactly what do you want to do, what you have posted is not a question. Please post the question as asked.P(x)= x^4-4x^3+10x^2+12x-39, using synthetic division given 2+3i is a zero of function

CB

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- #3

- Feb 7, 2012

- 2,818

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 $P(x)$.P(x)= x^4-4x^3+10x^2+12x-39, using synthetic division given 2+3i is a zero of function

- Jan 29, 2012

- 1,159

[tex](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[/tex] divides into [tex]x^4- 4x^3+ 10x^2+ 12x- 39[/tex] without remainder but synthetic division by x- (2+ 3i) is

2+3i|1_____-4_______10_________12_______-39

__________2+3i_____-13_______-6+9i_______+39

____1____-2+3i______-3________6+9i________0

or [tex]x^3+ (-2+3i)x^2- 3x+ (6+ 9i)[/tex]