a²-216 6) 64x³-27 Name Date Period n=x Zeros2 Kuta Software – Infinite Algebra 1 Factoring Trinomials (a > 1) Factor each completely. a b C 1) 3p²-2p-5 2) 2n²+3n-9 3) 3n²-8n+4 5) 2v²+11v+5 6) 2n²+5n+2 7) 7a²+53a +28 8) 9k²+66k+21 9) 15n²-27n-6 10) 5x²-18x+9 11) 4n²-15n-25 12) 4x²-35x+49 13) 4n²-17n+4 14) 6x²+7x-49 15) 6x²+37x+6 16)-6a²-25a-25 17) 6n²+5n-6 18) 16b²+60b-100
The Correct Answer and Explanation is:
1) 3p2−2p−53p^2 – 2p – 53p2−2p−5
We look for two numbers that multiply to 3⋅(−5)=−153 \cdot (-5) = -153⋅(−5)=−15 and add to −2-2−2.
Those numbers are 3 and -5.
Now rewrite the middle term: 3p2+3p−5p−53p^2 + 3p – 5p – 53p2+3p−5p−5
Group terms: (3p2+3p)−(5p+5)(3p^2 + 3p) – (5p + 5)(3p2+3p)−(5p+5)
Factor each group: 3p(p+1)−5(p+1)3p(p + 1) -5(p + 1)3p(p+1)−5(p+1)
Factor the common binomial: (3p−5)(p+1)(3p – 5)(p + 1)(3p−5)(p+1)
✅ Answer: (3p−5)(p+1)(3p – 5)(p + 1)(3p−5)(p+1)
Textbook-style Explanation (approx. 300 words)
Factoring trinomials where the leading coefficient aaa is greater than 1 involves a strategy called factoring by grouping. This method begins by identifying two numbers that multiply to a⋅ca \cdot ca⋅c (the product of the first and last coefficients) and add to bbb, the middle coefficient.
Take the example 3p2−2p−53p^2 – 2p – 53p2−2p−5. Here, a=3a = 3a=3, b=−2b = -2b=−2, and c=−5c = -5c=−5. We first compute a⋅c=3⋅(−5)=−15a \cdot c = 3 \cdot (-5) = -15a⋅c=3⋅(−5)=−15. Our goal is to find two integers that multiply to -15 and add up to -2. These numbers are 3 and -5.
Next, we rewrite the middle term −2p-2p−2p using the two numbers we found: 3p2+3p−5p−53p^2 + 3p – 5p – 53p2+3p−5p−5
Now, we group the terms in pairs: (3p2+3p)−(5p+5)(3p^2 + 3p) – (5p + 5)(3p2+3p)−(5p+5)
Factor out the greatest common factor (GCF) from each group: 3p(p+1)−5(p+1)3p(p + 1) – 5(p + 1)3p(p+1)−5(p+1)
We now see a common binomial factor: (p+1)(p + 1)(p+1). Factor it out: (3p−5)(p+1)(3p – 5)(p + 1)(3p−5)(p+1)
This is the completely factored form of the trinomial. The solution shows how the middle term can be broken into two terms to make factoring easier, emphasizing the importance of recognizing factor pairs of the product acacac. Practicing this process builds fluency and confidence in factoring more complex quadratic expressions.
