$(x+a)(x+b+ci)(x+b-ci)$

Algebra Level 2

Suppose that $a, b, c$ are positive reals such that $\begin{aligned} & x^3+7x^2+24x+18 \\ &= (x+a)(x+b+ci)(x+b-ci), \end{aligned}$ where $i$ is the imaginary number that satisfies $i^2=-1.$ What is the value of $a+b+c?$

7 9 8 6

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Mohd Naim Mohd Amin
May 21, 2014

Yo, x^3 + 7x^2 + 24x + 18

= x^3 + (x^2 + 6x^2) + (6x + 18x) + 18

= x^2(x+1) + 6x(x+1) + 18(x+1)

=(x+1)(x^2 + 6x + 18)

as (x+1)(x^2 + 6x + 18) = 0,

x^2 + 6x + 18 = 0 -----> solve by completing the square for this,

(x + 3 + 3i) ( x + 3 - 3i) = 0,

so (x+1)(x+3+3i)(x+3-3i) = (x+a)(x+b+ci)(x+b-ci)

a=1, b=c=3,

a+b+c = 1+3+3=7.....

thanks yo...

( x + a ) ( x + b + c i ) ( x + b − c i ) (x+a)(x+b+ci)(x+b-ci) ( x + a ) ( x + b + c i ) ( x + b − c i )

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$(x+a)(x+b+ci)(x+b-ci)$