Solving by factorisation

Factorisation rests on one small, powerful fact called the zero product rule: if two numbers multiply to give zero, at least one of them must itself be zero. There is no other way to get a product of exactly $0$.

where: $p$ and $q$ are any two real numbers (here, the two linear factors of the quadratic). To reach that factored form, CBSE uses the splitting the middle term technique. Two numbers $p$ and $q$ are found so that $p \times q = a\times c$ and $p+q=b$; the middle term $bx$ is then rewritten as $px+qx$, and the four resulting terms are grouped in pairs to pull out a common binomial factor.

where: $a$, $b$, $c$ are the coefficients of the standard quadratic, and $\alpha$, $\beta$ are its two roots. Once both roots are found, they can be sanity-checked against $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$ — if either check fails, an arithmetic slip crept in somewhere. Not every quadratic splits into nice integers. When no such $p,q$ exist as simple numbers, factorisation becomes impractical, and the quadratic formula (a separate technique) is used instead — factorisation is best thought of as the fast first attempt, not the universal tool. Factorisation is popular in exams precisely because it needs only integer arithmetic when the numbers cooperate — no square roots to simplify and no fractions to carry through several lines of working. Many textbook quadratics are deliberately constructed so that $ac$ splits into two friendly integer factors, which is exactly why this is usually the method worth trying first, before reaching for the quadratic formula. The skill is largely pattern recognition, and it improves with repetition: the more factor pairs you have tried mentally, the faster you spot which split works.

Solve $2x^2+5x-3=0$ by factorisation.

Step 1. Identify $a$, $b$, $c$. Here $a=2$, $b=5$, $c=-3$, so $ac = 2\times(-3)=-6$.

Step 2. Find two numbers that multiply to $-6$ and add to $5$. Try $6$ and $-1$: $6\times(-1)=-6$ and $6+(-1)=5$. These work.

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

Step 4. Group and factor each pair: $2x(x+3)-1(x+3)=0$.

Step 5. Factor out the common binomial: $(x+3)(2x-1)=0$.

Step 6. Apply the zero product rule. $x+3=0 \Rightarrow x=-3$, and $2x-1=0 \Rightarrow x=\dfrac{1}{2}$.

Roots: $x=-3$ or $x=\dfrac{1}{2}$.

Competitive stretch

Check the roots against the sum/product shortcut before moving on. Sum: $\alpha+\beta = -3+\dfrac{1}{2}=-\dfrac{5}{2}$, and $-\dfrac{b}{a}=-\dfrac{5}{2}$ — matches. Product: $\alpha\beta = -3\times\dfrac{1}{2}=-\dfrac{3}{2}$, and $\dfrac{c}{a}=\dfrac{-3}{2}$ — matches too. This 10-second check catches most factorisation slips before they cost marks, and it is exactly how the word problem below can be verified: for $x^2+x-90=0$ with roots $9$ and $-10$, sum $=-1=-b/a$ and product $=-90=c/a$, confirming $9$ and $10$ (after rejecting the negative root) are correct for the consecutive-integer problem whose product is $90$.

• Applying the zero product rule before the equation equals zero. $(\ldots)(\ldots)=0$ is required first; never split $(x-3)(x+2)=6$ directly into $x-3=0$ or $x+2=0$.
• Picking $p,q$ that satisfy only one condition. Always check both $pq=ac$ and $p+q=b$, not just the sum or only the product.
• Losing a sign while cancelling a common factor, e.g. going from $(2x-4)=0$ to $x=2$ is fine, but rushing the step can flip a sign — write out $2x=4$ before dividing.
• Not rejecting invalid roots in word problems. A negative length, a negative count of people, or (as in the consecutive-integer problem) a negative integer when the problem demands positive values must be discarded.
• Substituting the wrong number back to verify. After solving, always plug the root back into the original equation, not into an intermediate grouped form — a slip in an earlier step can otherwise go unnoticed.

Ask the tutor: Stuck on finding the right pair of numbers to split the middle term, or want to practise turning a word problem into a factorisable quadratic? Ask the AI tutor for a guided walkthrough.