If P(x)= x^3 +ax^2 +bx+c, with a, b and c real numbers. Roots of P(x) z+3i, z+9i and 2z−4, find ∣a+b+c∣. Note: z is complex number.


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Question Number 112478 by dw last updated on 08/Sep/20

$I f P (x) = x^{3} + {a x}^{2} + b x + c, w i t h a, b a n d c r e a l n u m b e r s .$ $R o o t s o f P (x) z + 3 i, z + 9 i a n d 2 z - 4, f i n d ∣ a + b + c ∣ .$ $N o t e : z i s c o m p l e x n u m b e r .$
	Answered by floor(10²Eta[1]) last updated on 08/Sep/20

	$we know that a cubic polynomial with$ $real coefficients can have :$ $3 real roots or 1 real root and 2 complex roots$ $because if a polynomial has a complex$ $root, the conjugate of this complex number$ $is also a root .$ $If z \in R \Rightarrow 2 z - 4 \in R, but if z + 3 i and z + 9 i$ $are the complex roots so z - 3 i and z - 9 i$ $also should be . But {that}^{'} s impossible$ $because we just have 3 roots .$ $So z have an imaginary part \Rightarrow 2 z - 4$ $is one of the complex roots .$ $z = u + vi$ $roots : [2 u - 4 + 2 vi], [u + (3 + v) i], [u + (9 + v) i]$ $1 case :$ $u + (9 + v) i is the real root :$ $\Rightarrow 9 + v = 0 ∴ v = - 9$ $so (2 u - 4 - 18 i) is the conjugate of (u - 6 i)$ $but this is impossible because the$ $imaginary parts are different .$ $2 case :$ $u + (3 + v) i is the real root :$ $\Rightarrow 3 + v = 0 ∴ v = - 3$ $so (2 u - 4 - 6 i) is the conjugate of (u + 6 i)$ $\Rightarrow 2 u - 4 = u ∴ u = 4$ $roots : (4), (4 + 6 i), (4 - 6 i)$ $\Rightarrow P (x) = (x - 4) (x - 4 - 6 i) (x - 4 + 6 i)$ $P (1) = 1 + a + b + c$ $= - 3 (- 3 - 6 i) (- 3 + 6 i) = - 3.45$ $\Rightarrow a + b + c = - 135 - 1 = - 136$ $∣ a + b + c ∣= 136 .$
	Commented by MJS_new last updated on 08/Sep/20

	$good job!$
	Commented by dw last updated on 09/Sep/20

	$T h a n k y o u S i r$
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