Factor each trinomial below. Please show your work and check your answer. (1 point each) x2 – 8x + 15 a2 – a – 20 a2 – 5a – 20 a2 + 12ab + 27b2 2a2 + 30a + 100
@amistre64
with any luck they play nice
What do you mean?
i was never good at grouping .... so i developed my own method
Oh, Sorry, I need to use "By the book" method. Thx anyway!
if we take a general setup:\[(ax+b)(mx+n)=(am)x^2+(an+bm)x+(bn)\] by multiplying the first and last coeffs we get:\[ambn\]which we would need to determine its factors \(an,~bm\) such that\[an+bm = an+bm\] as such, we can then factor it as\[(x+\frac{an}{ambn})(x+\frac{bn}{ambn})\] and reduce as needed
ugh ... got that last line a little convoluted :)\[(x+\frac{an}{am})(x+\frac{bm}{am})\]which reduces to\[(mx+n)(ax+b)~:~\text{ which was what we started with}\]
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