Unit 3 · Topic 05 · Algebra
Meera and Arjun raced to factorise x² + 7x + 12. Meera finished in one line. Arjun tried five different number pairs before he found the right one.
"You didn't guess faster," Meera said. "You just weren't checking a rule first."
She showed him the two questions that turn guessing into a two-step search.
"To factorise x² + 7x + 12," Meera said, "you need two numbers that MULTIPLY to give 12, the constant term, and ADD to give 7, the coefficient of x." Arjun started listing pairs that multiply to 12: 1 and 12, 2 and 6, 3 and 4. He checked each pair's sum: 13, 8, 7. The last pair worked.
"So x² + 7x + 12 becomes (x + 3)(x + 4)," he said, and checked by expanding: x² + 4x + 3x + 12 = x² + 7x + 12. It matched.
Meera then gave him x² − x − 12. "Same rule," she said. "Two numbers that multiply to −12 and add to −1." Arjun thought about it differently this time: since the product is negative, the two numbers must have opposite signs. Since the sum is only −1, a small negative number, the numbers can't be too far apart. He tried 3 and −4: product −12, sum −1. Both matched.
"x² − x − 12 factors as (x + 3)(x − 4)," he said, checking again by expanding: x² − 4x + 3x − 12 = x² − x − 12. Correct.
Then Meera made it harder: 2x² + 7x + 3, where the x² term has a coefficient other than 1. "Now you multiply the FIRST coefficient by the LAST term," she said. "2 times 3 is 6. Find two numbers that multiply to 6 and add to 7 — the middle coefficient. That's 1 and 6."
Arjun used those two numbers to SPLIT the middle term: 2x² + 1x + 6x + 3. Then he grouped in pairs, the same regrouping trick from before: (2x² + x) + (6x + 3) = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). He checked by expanding, and it matched.
"So splitting the middle term is really two rules stitched together," Arjun said. "Find the right two numbers using the product-and-sum test, split the middle term into those two pieces, then factor by grouping like we already know how to do."
"Exactly," said Meera. "The hard part was never the algebra. It was knowing which two numbers to hunt for — multiply to the FIRST TIMES LAST, add to the MIDDLE."
By the end of the session, Arjun wasn't listing five pairs any more. He was writing the product-and-sum test first, every time, before touching a bracket.
To factorise a quadratic ax² + bx + c, find two numbers that multiply to give a×c (the first coefficient times the last term) and add to give b (the middle coefficient). When a = 1, this simplifies to: multiply to c, add to b.
Once found, use those two numbers to split the middle term into two pieces, turning a three-term expression into a four-term one that can be factorised by grouping — exactly the technique from the regrouping topic.
The sign pattern is a useful shortcut: if c is positive, the two numbers have the SAME sign as b; if c is negative, the two numbers have OPPOSITE signs, and the larger one carries the sign of b.
This method always works for factorisable quadratics; when no pair of integers satisfies both conditions, the quadratic does not factor neatly over whole numbers.
Factorise x² + 8x + 15 and x² − 2x − 15 by finding the product-and-sum pair for each, then checking by expanding.
Factorise 3x² + 11x + 6 using the full first-times-last method, writing out every step: the target product, the pair found, the split, and the grouping.
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