Please help will fan and medal! what are the zeroes of y=x^2+14x+40?
I can solve this by factorization right?
standard form ax^2 +bx+c when a=1 as you have, then you want to find two numbers that multiply to give c, but add up to give b so list out the set of each 2 numbers that multiply to give 40 1 40 2 20 4 10 -1 -40 -2 -20 -4 -10 etc now all of the signs are positive so our multipliers will be positive so multiply to give 40, add to give 14 our set of multipliers will be 10 and 4 we know they are both positive hence (x+10)(x+4) so another example x^2 + 4x+4 multipliers of 4 1 4 2 2 -1 -4 -2 -2 everything is positive so the multipliers are positive multiply to give 4, but add to give 4 2 and 2 (x+2)(x+2)
try to use the same method for similar questions
Thank you so much!!!!! I will
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