Did you find any dificult in factoring the general trionomials?why?
Answers
Answer:
No. Since there is already a pattern in factoring trinomials that are not perfect square.
Step-by-step explanation:
The factors of a trinomial which are not perfect squares is equal to the product of the factors of the first and last terms.
Examples:
p²+ 5p + 6
v²+ 4v – 21
2q³ – 6q²– 36q
To factor p² + 5p + 6, factor the first term: p² = (p)(p). Factor the last term: 6 = (2)(3). Then, write the two as factors of the trinomial: (p + 2)(p + 3). Check if the sum of the factors is equal to the middle term which is 5. Thus, p² + 5p + 6 = (p + 2)(p + 3).
To factor v² + 4v – 21, factor the first term: v² = (v)(v). Factor the last term: -21 = (-3)(7). Then, write the two as factors of the trinomial: (v - 3)(v + 7). Check if the sum of the factors is equal to the middle term which is 4. Thus, v² + 4v – 21 = (v - 3)(v + 7).
To factor 2q³ – 6q² – 36q, divide the entire trinomial with 2q such that, 2q(q² - 3q - 18). Then, factor q² - 3q - 18 such that q² = (q)(q) and -18 = (3)(-6). Then write the two as factors of the trinomial: (q + 3)(q - 6). Check if the sum of the factors is equal to the middle term which is -3. Thus, 2q³ – 6q² – 36q = (2q)(q + 3)(q - 6).
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