Q.Solve the quadratic equation $x^2 - 5x + 6 = 0$ by factorization.

This equation factorizes because we can split $-5$ into two numbers whose product is $6$ (the constant term) — namely $-2$ and $-3$.

$x^2-2x-3x+6=0 \Rightarrow x(x-2)-3(x-2)=0 \Rightarrow (x-2)(x-3)=0$, so $x=2$ or $x=3$.

We need two numbers $p,q$ with $p\times q = 1\times 6 = 6$ and $p+q=-5$. Testing factor pairs of 6: $(-2)\times(-3)=6$ and $(-2)+(-3)=-5$. So $p=-2, q=-3$.

Split the middle term: $x^2 -2x -3x+6=0$.

Group in pairs: $(x^2-2x) + (-3x+6) = 0 \Rightarrow x(x-2) -3(x-2) = 0$.

Factor out the common binomial: $(x-2)(x-3) = 0$.

By the zero-product rule, $x-2=0$ or $x-3=0$, giving $x=2$ or $x=3$.

Check: $2^2-5(2)+6 = 4-10+6=0$ ✓. $3^2-5(3)+6=9-15+6=0$ ✓. Both roots verified by direct substitution.

Alternative — Sridharacharya's formula: here $a=1,b=-5,c=6$, so $D=(-5)^2-4(1)(6)=25-24=1$. Then $x=\dfrac{5\pm\sqrt{1}}{2}=\dfrac{5\pm1}{2}$, giving $x=3$ or $x=2$ — the same pair of roots, confirming the factorization answer.

A frequent error is choosing $p,q$ that add to $b$ but whose product is $c$ instead of $ac$ (they coincide only when $a=1$, which can mask the mistake in simple examples). Another common slip is a sign error when splitting the middle term — always check by re-expanding the two factors before finalizing the answer.