Express \( x^{2}+x \) in the form \( a x(x+2)+b(x+3)+c \)

To express \( x^{2} + x \) in the form \( a x(x + 2) + b(x + 3) + c \), we first expand the right-hand side: \[ a x(x + 2) + b(x + 3) + c = a x^2 + 2a x + b x + 3b + c \] Combining like terms, we have: \[ = a x^2 + (2a + b)x + (3b + c) \] Now, we want this expression to be equal to \( x^2 + x \). By equating coefficients, we get the following system of equations: 1. For \( x^2 \): \( a = 1 \) 2. For \( x^1 \): \( 2a + b = 1 \) 3. For the constant term: \( 3b + c = 0 \) Now, substitute \( a = 1 \) into the second equation: \[ 2(1) + b = 1 \implies 2 + b = 1 \implies b = 1 - 2 = -1 \] Next, substitute \( b = -1 \) into the third equation: \[ 3(-1) + c = 0 \implies -3 + c = 0 \implies c = 3 \] Now we have \( a = 1 \), \( b = -1 \), and \( c = 3 \). Thus, we can express \( x^{2} + x \) as: \[ x^{2} + x = 1 \cdot x(x + 2) - 1(x + 3) + 3 \] Putting it all together, we have: \[ x^{2} + x = x(x + 2) - (x + 3) + 3 \] In conclusion, the values of \( a \), \( b \), and \( c \) are: \[ a = 1, \quad b = -1, \quad c = 3 \]