Let f(x) = 2x^2 + 5x + 1. If we write f(x) as | vedclass

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(C) Given, $f(x) = 2x^2 + 5x + 1$. Also, $f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$. Expanding the right side: $f(x) = a(x^2 - x - 2) + b(x^2 -

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exactly one choice for each of $a, b, c$

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(C) Given, $f(x) = 2x^2 + 5x + 1$. Also, $f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$. Expanding the right side: $f(x) = a(x^2 - x - 2) + b(x^2 - 3x + 2) + c(x^2 - 1)$ $f(x) = (a+b+c)x^2 + (-a-3b)x + (-2a+2b-c)$. On equating the coefficients of $x^2$, $x$, and the constant term, we get a system of linear equations: $1) \; a + b + c = 2$ $2) \; -a - 3b = 5$ $3) \; -2a + 2b - c = 1$ Since we have $3$ linear equations in $3$ variables $(a, b, c)$ and the determinant of the coefficient matrix is non-zero, there exists a unique solution for $a, b, c$. Solving these, we get $a = -\frac{35}{4}$, $b = \frac{5}{4}$, and $c = \frac{38}{4}$. Thus, there is exactly one choice for each of $a, b, c$.

Let $f(x) = 2x^2 + 5x + 1$. If we write $f(x)$ as $f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$ for real numbers $a, b, c$, then:

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