Polynomials are Fun... :/

Let the zeroes of the polynomial be a-d, a, a+d respectively.

=> sum of the roots = 3a = 33 => a = 11

Next, sum of roots taken two at a time = 3a² − d² = 354 => d² = 9 => d = ±3

The roots are 8, 11, 14.

=> k = -8 * 11 * 14 = -1232

Let the roots be $a-d, a, a+d$ .

Now applying Vieta's formula,

$a-d+a+a+d=33$ $\Rightarrow a=11$

Also $(11-d)11+11(11+d)+(11-d)(11+d)= 354$ $\Rightarrow d^2=9$

Now $k=11(11-d)(11+d)$

$\Rightarrow k=-11(121-d^2)$ $= -11(121-9)$ $\huge \Rightarrow k=\boxed {-1232}$

Simple really. Since the roots are in A.P, we can express them as $r-d,r,r+d$ . The sum is $\frac { -b }{ a }$ which in this case is $33$ .

Adding up the roots would cancel the d's (that's what she said!) and becomes: $(r-d)+r+(r+d)=33$ and $r=11$

Now there's another useful concept of the Vieta's formula: if the roots of $f(x)=a{ x }^{ 3 }+b{ x }^{ 2 }+cx+d$ are $\alpha ,\beta ,\gamma$ , then $\alpha \beta +\beta \gamma +\alpha \gamma =\frac { b }{ a }$

Applying this concept, we substitute the roots: $(r)(11-d)+r(11+d)+(11-d)(11+d)=3{ 11 }^{ 2 }-{ d }^{ 2 }=354$

Solving this equation gives us $d=\pm 3$ . It doesn't matter which 3 we use because the result will still be in A.P.

The product of the roots in an ODD-DEGREE EQUATION is the constant divide by the leading term then multiply by $-1$ .

Substitution: $(11-3)(11)(11+3)=-k=-1232$