Factoring Algebraic Expressions Quiz — Secondary 4

Answer: 3x(2x + 3)

Always look for a common factor before anything else. Handle the numbers and the letters separately. For the numbers, the greatest common divisor of 6 and 9 is 3. For the letters, both terms contain x — the first has x² and the second has x, so the most you can pull out is x, the smaller power. The common factor is therefore 3x. Divide each term by it: 6x² ÷ 3x = 2x and 9x ÷ 3x = 3. So 6x² + 9x = 3x(2x + 3). The other options are incomplete factorings: 3(2x² + 3x) still has an x to take out, and x(6x + 9) still has a 3. "Completely" means nothing is left inside the bracket to pull out. Check by expanding: 3x × 2x = 6x² and 3x × 3 = 9x.

Answer: (x − 7)(x + 7)

Two signals tell you this is a difference of squares: there are only two terms, they are subtracted, and each one is a perfect square. Here x² is the square of x and 49 is the square of 7. The identity is a² − b² = (a − b)(a + b), so with a = x and b = 7 you get (x − 7)(x + 7). Expand to check: x² + 7x − 7x − 49, and the two middle terms cancel, leaving x² − 49. That cancellation is exactly why the pattern works. Careful with (x − 7)²: that equals x² − 14x + 49, which is a different expression altogether. And note that a sum of squares such as x² + 49 genuinely cannot be factored over the real numbers — only the difference can.

Answer: (x + 3)(x + 4)

For a trinomial x² + bx + c where the coefficient of x² is 1, look for two numbers whose product is c and whose sum is b. Here you need two numbers that multiply to 12 and add to 7. List the pairs that multiply to 12: 1 and 12 (sum 13), 2 and 6 (sum 8), 3 and 4 (sum 7). The last pair works, so x² + 7x + 12 = (x + 3)(x + 4). Expand to be sure: x² + 4x + 3x + 12 = x² + 7x + 12. Correct. Note that (x − 3)(x − 4) would give x² − 7x + 12 — the product of the two negatives is still positive, so the constant stays +12, but the middle term flips sign. That is the detail to watch when the signs change.

Answer: (x − 7)(x + 2)

You need two numbers whose product is −14 and whose sum is −5. Because the product is negative, the two numbers must have opposite signs; and because the sum is negative, the one with the bigger absolute value must be the negative one. The pairs multiplying to 14 are 1 and 14, or 2 and 7. With signs: −7 and +2 give a product of −14 and a sum of −5. That is the pair, so x² − 5x − 14 = (x − 7)(x + 2). Expand to confirm: x² + 2x − 7x − 14 = x² − 5x − 14. Correct. The option (x + 7)(x − 2) gives +5x in the middle, the right size but the wrong sign — the classic near-miss on this type.

Answer: 8

Expand the factored form and compare. Multiply term by term: (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Line that up against x² + bx + 15 and the middle coefficients must match, so b = 8. There is a shortcut: in a trinomial with leading coefficient 1, the middle coefficient is always the sum of the two numbers in the brackets, and the constant is always their product. Here 3 + 5 = 8 and 3 × 5 = 15, which is exactly what the expansion produced. Knowing that lets you factor and expand in your head instead of writing four terms each time.

Answer: (2x + 1)(x + 3)

When the coefficient of x² is not 1, use the product-sum method on a × c. Here a = 2 and c = 3, so a × c = 6, and you need two numbers with product 6 and sum 7 (the middle coefficient): those are 1 and 6. Split the middle term accordingly: 2x² + x + 6x + 3. Now group in pairs and factor each pair: x(2x + 1) + 3(2x + 1). Both groups share the bracket (2x + 1), so pull it out: (2x + 1)(x + 3). Expand to check: 2x² + 6x + x + 3 = 2x² + 7x + 3. Correct. The near-miss (2x + 3)(x + 1) gives 2x² + 5x + 3 — right first and last term, wrong middle — which is why you should always expand before moving on.

Answer: (x − 6)²

Test whether this is a perfect square trinomial. The first term x² is the square of x, and the last term 36 is the square of 6. For the pattern to hold, the middle term must be twice the product of those two roots: 2 × x × 6 = 12x. It is there, with a minus sign, which matches a² − 2ab + b² = (a − b)². So the factoring is (x − 6)². Expand to verify: (x − 6)(x − 6) = x² − 6x − 6x + 36 = x² − 12x + 36. Correct. The option (x − 4)(x − 9) also gives 36 as a product, but 4 + 9 = 13, not 12, so it fails. And (x − 6)(x + 6) is the difference of squares x² − 36, which has no middle term at all.

Answer: (x + 2)(x² + 3)

Four terms with no factor common to all of them is the signal to try grouping. Cut the expression in two halves: (x³ + 2x²) + (3x + 6). Factor each half on its own. From the first, pull out x²: x²(x + 2). From the second, pull out 3: 3(x + 2). The two halves now share the identical bracket (x + 2), and that is what makes the method work — if the brackets had come out different, you would need to regroup the terms in another order. Pull the common bracket out front: (x + 2)(x² + 3). Expand to check: x³ + 3x + 2x² + 6, which is the original once reordered. The near-miss (x + 2)(x² − 3) differs by a single sign and expands to x³ + 2x² − 3x − 6, so the last two terms come out negative: always expand before you commit. And note that x² + 3 cannot be factored further over the reals, since it is a sum of squares.

Answer: 5

Factor the left side first. You need two numbers with product 20 and sum −9: both must be negative, and −4 and −5 work, since (−4)(−5) = 20 and −4 + (−5) = −9. So the equation becomes (x − 4)(x − 5) = 0. Now apply the zero product rule: if a product of two factors equals zero, at least one factor must be zero. That gives x − 4 = 0 or x − 5 = 0, so x = 4 or x = 5. The larger solution is 5. Check it: 5² − 9(5) + 20 = 25 − 45 + 20 = 0. Correct. One warning about the zero product rule: it only works against zero. If the equation had read (x − 4)(x − 5) = 6, you could not conclude anything about the factors — you would have to expand, bring everything to one side, and factor again.

Answer: (2x − 5)(2x + 5)

This is still a difference of squares, but the first square has a coefficient. Ask what squares to give 4x²: since 2² = 4 and (x)² = x², the answer is 2x, because (2x)² = 4x². And 25 is the square of 5. Apply a² − b² = (a − b)(a + b) with a = 2x and b = 5: (2x − 5)(2x + 5). Expand to check: 4x² + 10x − 10x − 25 = 4x² − 25. Correct. The option 4(x − 5)(x + 5) expands to 4x² − 100, which is not the same thing — factoring out the 4 here is wrong because the 4 does not divide 25. The lesson: take the square root of each whole term, coefficient included.

Answer: 3(x − 2)(x + 2)

This question is really about order of operations in factoring: common factor first, special patterns second. Both terms are divisible by 3, so start there: 3x² − 12 = 3(x² − 4). Do not stop — the bracket is a difference of squares, since x² is the square of x and 4 is the square of 2. Factor it too: x² − 4 = (x − 2)(x + 2). The complete factoring is 3(x − 2)(x + 2). Expand to verify: (x − 2)(x + 2) = x² − 4, times 3 gives 3x² − 12. Now the two options you have to reject carefully, because neither is a false statement. 3(x² − 4) really does equal 3x² − 12 — it is simply the right idea stopped one step early, with a difference of squares left inside. And (3x − 6)(x + 2) also expands back to 3x² − 12 — it is a genuine factoring, but the first factor still has a 3 to pull out, so it is not complete either. That is exactly why the question says completely: a factoring is finished only when no factor can be broken down any further. The last option takes the square root of 12 as if it were 4, which it is not. Notice too how much easier the work becomes when the 3 comes out first: applying the difference of squares directly to 3x² − 12 would have left you with irrational roots.

Answer: 3 cm

The area of a rectangle is width times length, so A = x(x + 5) = x² + 5x. Notice that the factored form x(x + 5) is the picture itself — factoring a quadratic is literally finding the side lengths of a rectangle with that area. Set the area equal to 24: x² + 5x = 24. Bring everything to one side so you can use the zero product rule: x² + 5x − 24 = 0. Factor it — you need two numbers with product −24 and sum 5, which are 8 and −3: (x + 8)(x − 3) = 0. So x = −8 or x = 3. Now think about the situation, not just the algebra: a width cannot be negative, so reject −8 and keep x = 3 cm. Check: width 3, length 8, area 3 × 8 = 24 cm². Correct. Discarding the solution that makes no physical sense is part of the answer, not an afterthought.