Answer :
To write the equation of the quartic polynomial f(x)f(x) in standard form given its zeros, we'll use the factored form of the polynomial and then expand it.
The zeros given are −4,−3,11,−4,−3,11, and 2222. Since f(x)f(x) is a quartic function, we have:
f(x)=a(x+4)(x+3)(x−11)(x−22)f(x)=a(x+4)(x+3)(x−11)(x−22)
Given that the leading coefficient of f(x)f(x) is 1111, we substitute this value for aa:
f(x)=11(x+4)(x+3)(x−11)(x−22)f(x)=11(x+4)(x+3)(x−11)(x−22)
Expanding this expression, we get:
f(x)=11(x2+7x+12)(x2−33x+242)f(x)=11(x2+7x+12)(x2−33x+242)
f(x)=11(x4−33x3+242x2+7x3−231x2+1694x+12x2−396x+2904)f(x)=11(x4−33x3+242x2+7x3−231x2+1694x+12x2−396x+2904)
f(x)=11x4−363x3+2662x2+77x3−8085x2+18634x+132x2−4356x+31944f(x)=11x4−363x3+2662x2+77x3−8085x2+18634x+132x2−4356x+31944
f(x)=11x4−286x3−5409x2+14278x+31944f(x)=11x4−286x3−5409x2+14278x+31944
So, the equation of the quartic polynomial f(x)f(x) in standard form is:
f(x)=11x4−286x3−5409x2+14278x+31944f(x)=11x4−286x3−5409x2+14278x+31944
