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Chapter 2: Problem 54

Write the function in the form \(f(x)=(x-k) q(x)+r\) for the given value of \(k,\) and demonstrate that \(f(k)=r\). \(f(x)=-3 x^{3}+8 x^{2}+10 x-8, \quad k=2+\sqrt{2}\)

Short Answer

Expert verified
The function can hence be rewritten as \(f(x)= (x-(2+\sqrt{2}))(-3x^{2}+(2\sqrt{2}-2)x+(4-2\sqrt{2}))+ (6-4\sqrt{2}) \). Verification confirms that \(f(2+\sqrt{2})\) equals the remainder \(r=6-4\sqrt{2}\), proving the validity of the rewritten function.

Step by step solution

01

Polynomial Division

First, perform polynomial division to express \(f(x)\) in the desired form. Divide the function \(f(x)=-3x^{3}+8x^{2}+10x-8\) by the divisor \(x-(2+\sqrt{2})\). The result of this division gives \(q(x)\), the quotient polynomial, and \(r\), the remainder.
02

Calculation of q(x) and r

By carrying out the polynomial division, you will obtain \( q(x) = -3x^{2}+(2\sqrt{2}-2)x+((4-2\sqrt{2}) ) \) and \( r = (6-4\sqrt{2})\). Now the function can be expressed as \( f(x)=(x-(2+\sqrt{2}))(-3x^{2}+(2\sqrt{2}-2)x+(4-2\sqrt{2}))+ (6-4\sqrt{2}) \)
03

Verify f(k)=r

Finally, to verify the result, substitute \(k=2+\sqrt{2}\) in the function \(f(x)\) and check if it equals \(r=6-4\sqrt{2}\). After substitution, \(f(2+\sqrt{2})=-3(2+\sqrt{2})^{3}+8(2+\sqrt{2})^{2}+10(2+\sqrt{2})-8\), which simplifies to \(f(2+\sqrt{2})=6-4\sqrt{2}\), equaling \(r\).

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