Put a little more thought into this.

Algebra Level 3

f ( x ) = x 4 − 8 x 3 + 18 x 2 − 8 x + 2 \large f(x)= x^4 - 8x^3 + 18x^2 - 8x +2

Evaluate the value of f ( 2 + 3 ) f(2+\sqrt 3) for the function above.


The answer is 1.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Chew-Seong Cheong
Oct 16, 2016

f ( x ) = x 4 − 8 x 3 + 18 x 2 − 8 x + 2 = x 4 − 8 x 3 + 24 x 2 − 32 x + 16 − 6 x 2 + 24 x − 14 = ( x − 2 ) 4 − 6 ( x 2 − 4 x + 4 ) + 10 = ( x − 2 ) 4 − 6 ( x − 2 ) 2 + 10 ⟹ f ( 2 + 3 ) = ( 3 ) 4 − 6 ( 3 ) 2 + 10 = 9 − 18 + 10 = 1 \begin{aligned} f(x) & = x^4-8x^3+18x^2-8x+2 \\ & = x^4-8x^3+24x^2-32x+16-6x^2+24x-14 \\ & = (x-2)^4 - 6(x^2-4x+4) + 10 \\ & = (x-2)^4 - 6(x-2)^2 + 10 \\ \implies f(2+\sqrt 3) & = (\sqrt 3)^4-6(\sqrt 3)^2+10 \\ & = 9 - 18 + 10 \\ & = \boxed{1} \end{aligned}

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+1. The most elegant method. :)

Tapas Mazumdar - 4 years, 8 months ago
Zee Ell
Oct 15, 2016

f ( x ) = x 4 − 8 x 3 + 18 x 2 − 8 x + 2 = ( x 4 − 8 x 3 + 17 x 2 ) + ( x 2 − 8 x + 17 ) − 15 = f(x) = x^4 - 8x^3 + 18 x^2 - 8x + 2 = (x^4 - 8x^3 + 17 x^2) + (x^2 - 8x + 17) - 15 =

= ( x 2 + 1 ) ( x 2 − 8 x + 17 ) − 15 = (x^2 + 1) (x^2 - 8x+ 17) - 15

Since

( 2 + 3 ) 2 = 7 + 4 3 (2+ \sqrt {3} )^2 = 7 + 4 \sqrt {3}

therefore:

f ( 2 + 3 ) = ( 8 + 4 3 ) ( 7 + 4 3 − 16 − 8 3 + 17 ) − 15 = f(2+ \sqrt {3} ) = (8 + 4 \sqrt {3} )(7 + 4 \sqrt {3} - 16 - 8 \sqrt {3} + 17) - 15 =

= ( 8 + 4 3 ) ( 8 − 4 3 ) − 15 = 64 − 48 − 15 = 1 = (8 + 4 \sqrt {3} )(8 - 4 \sqrt {3}) - 15 = 64 - 48 - 15 = \boxed {1}

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