\( 612 a ^ { 2 } + 23 a b + 10 b ^ { 2 } \)

The polynomial expression you've provided is: \[ 612a^2 + 23ab + 10b^2 \] To determine whether this quadratic expression in two variables can be factored into the product of two binomials with integer coefficients, let's attempt to factor it. ### General Form for Factoring Assume the polynomial can be factored as: \[ (m a + n b)(p a + q b) = 612a^2 + 23ab + 10b^2 \] Expanding the left side: \[ m p a^2 + (m q + n p) ab + n q b^2 = 612a^2 + 23ab + 10b^2 \] This gives us a system of equations based on the coefficients: 1. **For \(a^2\) term:** \[ m p = 612 \] 2. **For \(ab\) term:** \[ m q + n p = 23 \] 3. **For \(b^2\) term:** \[ n q = 10 \] ### Exploring Possible Factors Let's explore possible integer pairs \((n, q)\) that satisfy \(n q = 10\): - \((1, 10)\) - \((2, 5)\) - \((5, 2)\) - \((10, 1)\) We'll test each pair to see if we can find corresponding integers \(m\) and \(p\) that satisfy all equations. #### Case 1: \(n = 2\), \(q = 5\) 1. From \(n q = 10\): \[ 2 \times 5 = 10 \quad \text{(satisfies)} \] 2. From \(m q + n p = 23\): \[ 5m + 2p = 23 \] 3. From \(m p = 612\), solve for \(p = \frac{612}{m}\) and substitute into the second equation: \[ 5m + 2\left(\frac{612}{m}\right) = 23 \] Multiply through by \(m\) to eliminate the fraction: \[ 5m^2 + 1224 = 23m \] Rearranging: \[ 5m^2 - 23m + 1224 = 0 \] The discriminant (\(D\)) of this quadratic equation is: \[ D = (-23)^2 - 4 \times 5 \times 1224 = 529 - 24480 = -23951 \] Since \(D < 0\), there are no real solutions for \(m\) and \(p\) in this case. #### Case 2: \(n = 5\), \(q = 2\) Following a similar process, you'll find that no integer solutions satisfy all equations. #### Case 3 & 4: \(n = 1, q = 10\) and \(n = 10, q = 1\) Both cases result in quadratic equations with negative discriminants, indicating no real (and certainly no integer) solutions. ### Conclusion Since none of the possible integer pairs \((n, q)\) lead to integer values of \(m\) and \(p\) that satisfy all three equations, **the polynomial \(612a^2 + 23ab + 10b^2\) cannot be factored into a product of two binomials with integer coefficients**. Furthermore, even if we consider real coefficients, the negative discriminant implies that the polynomial is **irreducible over the real numbers**. If complex coefficients are allowed, it can be factored using complex numbers, but such factorization typically exceeds standard factoring requirements. **Final Answer:** The expression \(612a^2 + 23ab + 10b^2\) cannot be factored into a product of two binomials with real (or integer) coefficients. It is irreducible over the real numbers.