Q.Solve $2x^2 + 3x - 2 = 0$ by factorization.

Here $a=2, c=-2$, so $ac=-4$; we need two numbers with product $-4$ and sum $3$ — those are $4$ and $-1$.

$2x^2+4x-x-2=0 \Rightarrow 2x(x+2)-1(x+2)=0 \Rightarrow (x+2)(2x-1)=0$, so $x=-2$ or $x=\tfrac12$.

General form: $a=2, b=3, c=-2$. We need $p,q$ with $p \times q = ac = 2 \times(-2) = -4$ and $p+q=b=3$. Testing: $4 \times (-1) = -4$ and $4+(-1)=3$. So $p=4,q=-1$.

Split the middle term: $2x^2+4x-x-2=0$.

Group: $(2x^2+4x)+(-x-2)=0 \Rightarrow 2x(x+2)-1(x+2)=0$.

Factor out $(x+2)$: $(x+2)(2x-1)=0$.

By the zero-product rule: $x+2=0 \Rightarrow x=-2$, or $2x-1=0 \Rightarrow x=\dfrac{1}{2}$.

Check: $2(-2)^2+3(-2)-2=8-6-2=0$ ✓. $2\left(\tfrac12\right)^2+3\left(\tfrac12\right)-2=\tfrac12+\tfrac32-2=0$ ✓.

Alternative — Sridharacharya's formula: $D=3^2-4(2)(-2)=9+16=25$. $x=\dfrac{-3\pm\sqrt{25}}{4}=\dfrac{-3\pm5}{4}$, giving $x=\dfrac{2}{4}=\dfrac12$ or $x=\dfrac{-8}{4}=-2$ — matching the factorization result exactly.

When $a \neq 1$, students often forget to multiply $a$ and $c$ before searching for the split numbers — searching for numbers that multiply to $c$ alone (here $-2$) instead of $ac$ (here $-4$) gives a wrong or impossible split. Also, after factoring out $(x+2)$, remember the leftover factor is $2x-1$, not $x-1$ — losing track of the coefficient in the grouped term is a common slip.