➕ Algebra 1 · Factoring
Factoring Trinomials flashcards
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All 17 flashcards at a glance
Card 1
To factor x2 + bx + c, which two numbers do you look for?
Two integers p and q with pq = c and p + q = b. Then x2 + bx + c = (x + p)(x + q). Take out any GCF first.
Card 2
Factor x2 + 10x + 21.
(x + 3)(x + 7). The pairs with product 21 are 1 · 21 (sum 22) and 3 · 7 (sum 10). Check: x2 + 7x + 3x + 21.
Card 3
Factor x2 − 11x + 24.
(x − 3)(x − 8). Both numbers are negative: (−3)(−8) = 24 and −3 + (−8) = −11.
Card 4
Factor x2 + x − 12.
(x + 4)(x − 3). The numbers have opposite signs, the larger is positive, and 4 + (−3) = 1 with 4 · (−3) = −12.
Card 5
Factor x2 − 5x − 24.
(x − 8)(x + 3). Opposite signs, the larger is negative: (−8)(3) = −24 and −8 + 3 = −5.
Card 6
If c is positive, what do you know about p and q in (x + p)(x + q)?
They have the same sign, and that sign is the sign of b. (If b = 0, no pair works, because two numbers with the same sign cannot add to 0, so the trinomial is prime, like x2 + 49.)
Card 7
If c is negative, what do you know about p and q?
They have opposite signs. If b ≠ 0, the one that is larger in size has the same sign as b. If b = 0, they are opposites, like 8 and −8.
Card 8
What do you do first, before any other step, when factoring a trinomial?
Take out the GCF (a negative GCF if the leading coefficient is negative), and keep it in the final answer.
Card 9
Factor 2x2 + 10x + 12 completely.
2(x + 2)(x + 3). The GCF is 2, which leaves x2 + 5x + 6. Then 2 · 3 = 6 and 2 + 3 = 5.
Card 10
In this lesson’s SPLICE trick for ax2 + bx + c, what do the letters stand for?
Strip out the GCF. Product: compute ac. Look for two integers with product ac and sum b. Insert them to split the middle term. Common factors of each pair, then the common binomial. Expand to check.
Card 11
Is x2 + 8x + 10 factorable using integers?
No, it is prime. The positive pairs with product 10 are 1 · 10 (sum 11) and 2 · 5 (sum 7), and negative pairs have negative sums. No pair adds to 8.
Card 12
How can you recognize a perfect square trinomial?
The first and last terms are perfect squares, and the middle term is twice the product of their square roots (plus or minus). Both end terms being squares is not enough.
Card 13
Factor 9x2 − 24x + 16.
(3x − 4)2. The end terms are (3x)2 and 42, and the middle term is −2 · 3x · 4 = −24x.
Card 14
Factor 3x2 + 8x + 4.
(3x + 2)(x + 2). Here ac = 12, and the pair with sum 8 is 2 and 6. Then 3x2 + 2x + 6x + 4 = x(3x + 2) + 2(3x + 2).
Card 15
Factor x2 − 64.
(x − 8)(x + 8). There is no middle term, so b = 0 and the two numbers add to 0. With product −64, they are 8 and −8. This is a difference of squares.
Card 16
For (x − 2)(x + 9), which x-values make the product 0, and why?
x = 2 and x = −9. A product is 0 only if a factor is 0, so x − 2 = 0 or x + 9 = 0.
Card 17
Why should you always expand your answer when you factor?
Expanding must give back the original trinomial. A pair with the right product but wrong signs gives the wrong middle term, and expanding catches it. It does not tell you whether the answer is completely factored.