If the zeroes of (p(x)=x^3+ax^2+bx-12) are (1,3,-4), what is (a+b)?

Correct answer: -(9) / (-9)

The direct answer is option A: \(a+b=-13\). For a monic cubic with zeroes \(\alpha,\beta,\gamma\), \(x^3+ax^2+bx+c=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma\). Here the zeroes are 1, 3 and -4. Their sum is \(1+3-4=0\), so \(-a=0\), giving \(a=0\). Their pairwise-product sum is \(1\cdot3+3\cdot(-4)+(-4)\cdot1=3-12-4=-13\), so \(b=-13\). Therefore \(a+b=0+(-13)=-13\). Option A is correct because it gives -13. Option B gives 9, but neither the required coefficient calculation nor the result is 9. Option C gives -7, which comes from an incorrect combination of the roots. Option D gives 7 and is also based on an incorrect sign or product. The constant term checks the roots: \(-1\cdot3\cdot(-4)=-12\), matching the polynomial. Memory cue: for a monic cubic, the coefficient of \(x^2\) is the negative of the root sum, while the coefficient of \(x\) is the pairwise-product sum.