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When \(f(x) = ax^3 - 6x^2 + bx - 5\) is divided by x - 1, the remainder is -5. When f(x) is divided by x + 2, the remainder is -53. Find the ordered pair (a,b).

 May 14, 2019
 #1
avatar+5074 
+2

deleted... had f(1)=5 not -5

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 May 15, 2019
edited by Rom  May 15, 2019
 #2
avatar+22180 
+2

When \(f(x) = ax^3 - 6x^2 + bx - 5\) is divided by \(x - 1\), the remainder is \(-5\).
When \(f(x)\) is divided by \(x + 2\), the remainder is \(-53\).
Find the ordered pair \((a,b)\).

 

In general:

If a polynomial \(f(x)\) is divided by \(x-a\) , the remainder is the constant \(f(a)\) , and \(f(x)=q(x)·(x-a)+f(a)\)

 

\(\begin{array}{|lrcll|} \hline (x-1),\ a= 1: & \mathbf{f(1)} &=& \mathbf{-5} \\ & a\cdot 1^3 - 6\cdot 1^2 + b\cdot 1 - 5 &=& -5 \\ & a - 6 + b - 5 &=& -5 \\ &\mathbf{ a + b } &=& \mathbf{6} \qquad (1) \\ & \mathbf{b} &=& \mathbf{6-a} \\ \hline (x+2),\ a= -2: & \mathbf{f(-2)} &=& \mathbf{-53} \\ & a\cdot(-2)^3 - 6\cdot (-2)^2 + b\cdot(-2) - 5 &=& -53 \\ & -8a - 24 -2b &=& -48 \quad | \quad : (-2) \\ & 4a +12 +b &=& 24 \\ & \mathbf{4a+b} &=& \mathbf{12} \qquad (2) \\ & 4a+b &=& 12 \quad | \quad b=6-a \\ & 4a+6-a &=& 12 \\ & 3a &=& 6 \\ &\mathbf{ a } &=& \mathbf{2} \\\\ & b &=& 6-a \\ & b &=& 6-2 \\ &\mathbf{ b } &=& \mathbf{4} \\ \hline \end{array}\)

 

\(\mathbf{(a,b) = (2,4)}\)

 

Check:

 

laugh

 
 May 15, 2019

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