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Write a quadratic equation whose roots are 2 and 5 and whose leading coefficient is 3
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They give you the zeros, so you can reconstruct the polynomial (x-2)(x-5)=0 To get a leading coefficient of 3, we multiply by 3: 3(x-2)(x-5)=0 Now you have to foil it out
\[3x^2-7x+10\] is that it?
n0
now, because if you plug in x=2 and x=5, you will not get 0
no*
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\[3x^2-11x+30\]
no, a little closer though
you messed up on the inner part of the foil
outer+inner rather
is that part 17x?
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no
foil (x-2)(x-5), then multiply by 3
first + outer + inner + last
oh wait i messed up its 21x
x*x + x*-5 + -2*x + -2*-5
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yea, 3x^2-21x+30
ok thanks
np
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