Factorization is an art (4)

Algebra Level 5

f ( x ) = x 3 + 21 x 2 + 147 x + 336 \large f(x)= x^3+21x^2+147x+336 Let f : R R f: \mathbb {R \to R} be defined as above.

If the real root of the equation f ( f ( f ( f ( f ( x ) ) ) ) ) 2016 times = 0 \underbrace{f(f(f(\dots f(f(x))\dots)))}_{2016\ \text{times}} =0 is of the form a + a 1 / 3 b \large -a+a^{1/ {3^b}} , where a a and b b are integer , find a + b a+b .


The answer is 2023.

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1 solution

Tommy Li
Jul 4, 2016

f ( x ) = x 3 + 21 x 2 + 147 x + 336 f(x)= x^3+21x^2+147x+336

f ( x ) = ( x + 7 ) 3 7 f(x)=(x+7)^{3}-7

f ( f ( x ) ) = ( ( x + 7 ) 3 7 + 7 ) 3 7 f(f(x))=((x+7)^{3}-7+7)^{3}-7

f ( f ( x ) ) = ( x + 7 ) 3 2 7 f(f(x))=(x+7)^{3^{2}}-7

f ( f ( f ( f ( f ( x ) ) ) ) ) 2016 times = ( x + 7 ) 3 2016 7 \large \Rightarrow \underbrace{f(f(f(\dots f(f(x))\dots)))}_{2016\ \text{times}} = (x+7)^{3^{2016}}-7

f ( f ( f ( f ( f ( x ) ) ) ) ) 2016 times = 0 \large \underbrace{f(f(f(\dots f(f(x))\dots)))}_{2016\ \text{times}} =0

( x + 7 ) 3 2016 7 = 0 \large (x+7)^{3^{2016}}-7 =0

x = 7 + 7 1 3 2016 \large \Rightarrow x = -7+7^{\dfrac{1}{3^{2016}}}

a + b = 2016 + 7 = 2023 \large \Rightarrow a+b =2016+7=2023

f ( f ( f ( f ( f ( x ) ) ) ) ) n times = ( x + 7 ) 3 n 7 \large \Rightarrow \underbrace{f(f(f(\dots f(f(x))\dots)))}_{n \ \text{times}} = (x+7)^{3^{n}}-7

You need to prove that the equation above holds true for ALL positive integers first. Hint : induction.

Pi Han Goh - 4 years, 11 months ago

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