➕ Algebra 1 · Factoring
Factoring by Grouping: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
x2(x + 5) + 3(x + 5) is already written as a product of factors.
✗ False. It is a sum of two terms. Each term is a product, but the whole expression is a sum. Pull out the common binomial (x + 5) to get the product (x + 5)(x2 + 3).
The binomials (x − 2) and (2 − x) are opposites, so one pair can be rewritten with a −1 taken out to make them match.
✓ True. 2 − x = −(x − 2), so taking out a −1 turns (2 − x) into (x − 2), with the sign moved out front.
x3 − 7x2 − 3x + 21 = x2(x − 7) + 3(x − 7).
✗ False. The second pair −3x + 21 equals −3(x − 7), not 3(x − 7). The correct step is x2(x − 7) − 3(x − 7). Test with x = 2: the original is 8 − 28 − 6 + 21 = −5, but x2(x − 7) + 3(x − 7) is 4(−5) + 3(−5) = −35.
Every four-term polynomial can be factored by grouping.
✗ False. Grouping needs matching binomials. In x3 + x2 + 3x + 1 the pairs give x2(x + 1) and (3x + 1), which do not match, and rearranging to x(x2 + 3) + (x2 + 1) does not match either (nor does the third way of pairing). Grouping fails here, but that alone does not prove the polynomial is prime. It only shows that grouping does not apply.
5x2 + 16x + 3 = 5x2 + 15x + x + 3, and grouping the right side gives (x + 3)(5x + 1).
✓ True. Here a × c = 15, and 15 × 1 = 15 and 15 + 1 = 16. Group: 5x(x + 3) + 1(x + 3) = (x + 3)(5x + 1). Check: (x + 3)(5x + 1) = 5x2 + x + 15x + 3.
To split the middle term of ax2 + bx + c, find two integers whose sum is a × c and whose product is b.
✗ False. It is the other way around: the product is a × c and the sum is b. For 3x2 + 8x + 4 the product is 12 and the sum is 8, which gives 2 and 6.
Reordering the terms of a polynomial does not change its value, so you can rearrange them before pairing, as long as each sign stays with its term.
✓ True. Addition can be done in any order. For example, 4 − 7x + 3x2 and 3x2 − 7x + 4 are equal for every x (at x = 2 both are 2).
In y3 + 4y2 + y + 4, the second pair y + 4 has GCF 1, and the factored form is (y + 4)(y2 + 1).
✓ True. y2(y + 4) + 1(y + 4) = (y + 4)(y2 + 1). Check: (y + 4)(y2 + 1) = y3 + y + 4y2 + 4.
Factoring by grouping only works on polynomials with one variable.
✗ False. It also works with several variables: xy + 3x + 4y + 12 = x(y + 3) + 4(y + 3) = (y + 3)(x + 4). Check: (y + 3)(x + 4) = xy + 4y + 3x + 12.
For 3x3 + 21x2 + 6x + 42, you can take out the GCF 3 first, leaving x3 + 7x2 + 2x + 14 inside the parentheses.
✓ True. Dividing each term by 3 gives x3 + 7x2 + 2x + 14. Then x2(x + 7) + 2(x + 7) = (x + 7)(x2 + 2), so the full answer is 3(x + 7)(x2 + 2).
x3 + 2x2 − 5x − 10 = (x + 2)(x2 − 5).
✓ True. Group: x2(x + 2) − 5(x + 2) = (x + 2)(x2 − 5). Check: (x + 2)(x2 − 5) = x3 − 5x + 2x2 − 10.
x3 + 3x2 − 2x − 6 = (x + 3)(x2 + 2).
✗ False. The second pair −2x − 6 is −2(x + 3), so the answer is (x + 3)(x2 − 2). Expanding (x + 3)(x2 + 2) gives x3 + 2x + 3x2 + 6, which has +2x and +6 instead of −2x and −6.