F(x)= Ax^3+Bx^2+Cx+D
When x=1 f(1)=21
When X=21 f(21)31061 what is a,b,c and d t equals 21 If its not 0 1 or 7

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## 1 Answer

f(1)=21=a+b+c+d; d=21-(a+b+c)

f(21)=9261a+441b+21c+d=31061;

9261a+441b+21c+21-(a+b+c)=31061;

9260a+440b+20c=31040;

926a+44b+2c=3104.

463a+22b+c=1552.

a<4, so:

let a=3: 1389+22b+c=1552, 22b+c=163.

If c has to be positive, then b<8. If b cannot be 7, then let b=6:

132+c=163, c=163-132=31.

So, if a=3, b=6, c=31 and d=21-(a+b+c)=21-40=-19.

If c can be negative, let b=8, 176+c=163, c=-13.

So, if a=3, b=8, c=-13, then d=21-(a+b+c)=21-(-2)=23.

The question doesn't specify any other constraints so there are several solutions, including negative values.

by Top Rated User (1.0m points)

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