What are all the possible rational zeros for #y=21x^2+11x-40# and how do you find all zeros?

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Answer 1 · 2016-08-19T15:00:55

The Zeroes are, #x=-5/3, and, 8/7#.

To find the zeroes of #y=21x^2+11x-40#, we have to factorise it.

We observe that,

#21xx40=(7*3)xx(5*8)=(7*5)xx(3*8)=35xx24#

and, #35-24=11#.

#:. y=ul(21x^2+35x)-ul(24x-40)#

#=7x(3x+5)-8(3x+5)#

#=(3x+5)(7x-8)#

Hence, the Zeroes are, #x=-5/3, and, 8/7#.