2y^2-41y+165
What is the best way to do this. Like how can someone think of two numbers that multiply to 165 and at the same try them all out each time so it adds up to -41
Is this trial and error. If this was on the test, how would I do it if I dont have much time when the numbers are bigger its harder to work with and think of numbers. What do I do, how do you approach this, and what is the final answer. Lol lots of questions, thanks! :)
What is the best way to do this. Like how can someone think of two numbers that multiply to 165 and at the same try them all out each time so it adds up to -41
Is this trial and error. If this was on the test, how would I do it if I dont have much time when the numbers are bigger its harder to work with and think of numbers. What do I do, how do you approach this, and what is the final answer. Lol lots of questions, thanks! :)
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(y-15)(2y-11)
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They'd have to multiply to 2*165 or 330, if I'm not mistaken, which seems annoying and not necessarily possible. Use the quadratic formula:
x = [-b plus or minus (b^2 - 4ac)^0.5]/2a
x = {-(-41) plus or minus [(-41)^2 - 4*2*165]^0.5}/2*2
x = {41 plus or minus [1681 - 1320]^0.5}/4
x = {41 plus or minus [361]^0.5}/4
x = {41 plus or minus 19}/4
x = [-b plus or minus (b^2 - 4ac)^0.5]/2a
x = {-(-41) plus or minus [(-41)^2 - 4*2*165]^0.5}/2*2
x = {41 plus or minus [1681 - 1320]^0.5}/4
x = {41 plus or minus [361]^0.5}/4
x = {41 plus or minus 19}/4
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2y^2-41y+165 = (2y-31)(y-5)
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factor 165 hella. thats really tough sry.