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Algebra Level 2

Suppose f ( x ) = 3 x 3 − 14 x 2 + 15 x + 9 f(x)=3{ x }^{ 3 }-14{ x }^{ 2 }+15x+9 . Find the value of A + B + C + D A+B+C+D if another way to write f ( x ) f(x) is f ( x ) = A ( x − 2 ) 3 + B ( x − 2 ) 2 + C ( x − 2 ) + D f(x)=A{ (x-2) }^{ 3 }+B{ (x-2) }^{ 2 }+C(x-2)+D .


The answer is 9.

Kevin Tran by Kevin Tran , 19, USA

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2 solutions

Chew-Seong Cheong
Feb 27, 2018

f ( x ) = A ( x − 2 ) 3 + B ( x − 2 ) 2 + C ( x − 2 ) + D Putting x = 3 f ( 3 ) = A + B + C + D f ( 3 ) = 3 ( 3 3 ) − 14 ( 3 2 ) + 15 ( 3 ) + 9 = 9 \begin{aligned} f(x) & = A(x-2)^3+B(x-2)^2+C(x-2) + D & \small \color{#3D99F6} \text{Putting }x=3 \\ f(3) & = A + B + C + D \\ f(3) & = 3(3^3) - 14(3^2) + 15(3)+ 9 = 9 \end{aligned}

Therefore, A + B + C + D = 9 A+ B+C+D = \boxed{9} .

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Smart way to look at it!

Peter van der Linden - 3 years, 3 months ago

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Thanks. I knew there was always a trick.

Chew-Seong Cheong - 3 years, 3 months ago
Steven Chase
Feb 26, 2018

( x − 2 ) 2 = x 2 − 4 x + 4 ( x − 2 ) 3 = x 3 − 6 x 2 + 12 x − 8 \large{(x-2)^2 =x^2 - 4x + 4 \\ (x-2)^3 = x^3 - 6 x^2 + 12 x - 8}

Equate coefficients on powers of x x from both equations:

3 = A − 14 = − 6 A + B 15 = 12 A − 4 B + C 9 = − 8 A + 4 B − 2 C + D \large{3 = A \\ -14 = -6 A + B \\ 15 = 12 A - 4 B + C \\ 9 = -8 A + 4 B - 2 C + D}

Solving these equations results in:

A = 3 B = 4 C = − 5 D = 7 \large{A = 3 \\ B = 4 \\ C = -5 \\ D = 7}

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