Answer:
a = 2, b = -18, c = 40
Step-by-step explanation:
[tex]Given:\ Polynomial\ f(x)=ax^2+bx+c\ has\ zoroes\ 4\ aud\ 5[/tex]
[tex]and\ f(-1)=60[/tex]
[tex]Tofind:\ a,\ b,\ c[/tex]
[tex]Solution:\ As\ 4\ and\ 5\ are\ zeroes\ so\ f(4)=f(5)=0[/tex]
[tex]f(4)=a(4)^2+b(4)+c=0\Rightarrow16a+4b+c=0-(1)[/tex]
[tex]f(5)=a(5)^2+b(5)+c=0\Rightarrow25a+5b+c=0[/tex]
[tex]f(-1)=a(-1)^2+b(-1)+c=60\Rightarrowa-b+c=60-(3)[/tex]
[tex]subtracting\ (1)\ from\ (2)[/tex]
[tex]25a+5b+c-(16a+4b+c)=0\Rightarrow9a+b=0\Rightarrowb=-9a[/tex]
[tex]subtracting\ (3)\ from(2):[/tex]
[tex]25a+5b+c-(a-b+c)=0-60\Rightarrow24a+6b=-60[/tex]
[tex]putting\ b=-9a[/tex]
[tex]24a+6(-9a)=-60\Rightarrow24a-54a=-60\Rightarrow-30a=-60[/tex]
[tex]a=\frac{-60}{-30}=2[/tex] [tex]\Rightarrow[/tex] [tex]a=2[/tex]
[tex]b=-9a=-9\times2=-18[/tex]
[tex]c=-a+b\neq60=-2-18+60=-20+60=40[/tex]
I hope this helps you
:)