The expression \[\frac{d^2}{dx^2}\bigl[(2x-1)^4(1-x)^5\bigr] + \frac{d}{dx}\bigl[(2x+1)^4(3x^2-2)^2\bigr]\] is a polynomial of degree

\(9\);

\(8\);

\(7\);

less than \(7\).

Let’s observe that

multiplying two polynomials of degrees \(p\) and \(q\) gives a polynomial of degree \(p+q\), and

differentiating a polynomial of degree \(p\) gives a polynomial of degree \(p-1\).

From this we can quickly see that the expression in the question is a sum of two polynomials, each of degree 7.

Does this mean their sum has to be a polynomial of degree 7 too?

Yes, unless the leading terms of the two expressions cancel when you add them together.

So what are our leading terms here?

The expression \((2x-1)^4(1-x)^5\) gives first term \((2x)^4(-x)^5=-16x^9\), which differentiates twice to \[-9 \times 8 \times 16 x^7.\]

The leading term of the expression \((2x+1)^4(3x^2-2)^2\) is \((2x)^4(3x^2)^2=16 \times 9x^8\), which differentiates to \[16 \times 9 \times 8x^7.\]

So in fact the two leading terms here DO cancel, and we’re left with a polynomial of degree less than \(7\), so the answer is (d).