Solve for \( x \) : 1. \( (x+3)(x-1)=0 \) 2. \( -2(x+2)(x+3)=0 \) 3. \( 5 x(2 x+1)=0 \) 4. \( x^{2}+14 x+48=0 \) 5. \( 2 x^{2}-18=0 \)

**1. Solve:** \( (x+3)(x-1)=0 \) Since a product is zero, we set each factor equal to zero: - \( x+3=0 \) ⟹ \( x=-3 \) - \( x-1=0 \) ⟹ \( x=1 \) Thus, the solutions are: \( x=-3 \) and \( x=1 \) --- **2. Solve:** \( -2(x+2)(x+3)=0 \) The constant \(-2\) does not affect the roots. So we set: - \( x+2=0 \) ⟹ \( x=-2 \) - \( x+3=0 \) ⟹ \( x=-3 \) Thus, the solutions are: \( x=-2 \) and \( x=-3 \) --- **3. Solve:** \( 5x(2x+1)=0 \) Since \(5\) is nonzero, we have: - \( x=0 \) - \( 2x+1=0 \) ⟹ \( 2x=-1 \) ⟹ \( x=-\frac{1}{2} \) Thus, the solutions are: \( x=0 \) and \( x=-\frac{1}{2} \) --- **4. Solve:** \( x^{2}+14x+48=0 \) We factor the quadratic by searching for two numbers that multiply to \(48\) and add to \(14\): \( 6 \) and \( 8 \) satisfy these conditions. Factorization gives: \[ x^2 + 14x + 48 = (x+6)(x+8)=0 \] Set each factor equal to zero: - \( x+6=0 \) ⟹ \( x=-6 \) - \( x+8=0 \) ⟹ \( x=-8 \) Thus, the solutions are: \( x=-6 \) and \( x=-8 \) --- **5. Solve:** \( 2x^{2}-18=0 \) First, divide the entire equation by 2: \[ x^2 - 9 = 0 \] Recognize the difference of squares: \[ x^2 - 9 = (x-3)(x+3)=0 \] Set each factor equal to zero: - \( x-3=0 \) ⟹ \( x=3 \) - \( x+3=0 \) ⟹ \( x=-3 \) Thus, the solutions are: \( x=3 \) and \( x=-3 \)