frac{3 x^{2}+1}{x^{2}-6 x+8} is equal to | vedclass

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Question

(A) Given expression: $\frac{3 x^{2}+1}{x^{2}-6 x+8}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long divi

Vedclass · Accepted Answer

$3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}$

Vedclass · Answer

(A) Given expression: $\frac{3 x^{2}+1}{x^{2}-6 x+8}$.Since the degree of the numerator is equal to the degree of the denominator,we perform long division:$\frac{3 x^{2}+1}{x^{2}-6 x+8} = 3 + \frac{18 x - 23}{x^{2}-6 x+8}$.Now,we decompose the remainder using partial fractions:$\frac{18 x - 23}{(x-2)(x-4)} = \frac{A}{x-2} + \frac{B}{x-4}$.$18 x - 23 = A(x-4) + B(x-2)$.Setting $x=2$: $18(2) - 23 = A(2-4) \Rightarrow 36 - 23 = -2A \Rightarrow 13 = -2A \Rightarrow A = -\frac{13}{2}$.Setting $x=4$: $18(4) - 23 = B(4-2) \Rightarrow 72 - 23 = 2B \Rightarrow 49 = 2B \Rightarrow B = \frac{49}{2}$.Substituting these values back into the expression:$\frac{3 x^{2}+1}{x^{2}-6 x+8} = 3 - \frac{13}{2(x-2)} + \frac{49}{2(x-4)}$.This matches option $A$.

$\frac{3 x^{2}+1}{x^{2}-6 x+8}$ is equal to

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