(a) Differentiate \((x - 3)(x^{2} + 5)\) with respect to x. (b) If \((x + 1)^{2}\) is a factor of \(f(x) = x^{3} + ax^{2} + bx + 3\), where a and b are cons...

Answer Details

(a) Differentiate \((x-3)(x^2+5)\).

Expanding first: \((x-3)(x^2+5)=x^3+5x-3x^2-15=x^3-3x^2+5x-15.\)

Differentiating term by term:

\[\frac{dy}{dx}=3x^2-6x+5.\]

(The product rule gives the same result: \((1)(x^2+5)+(x-3)(2x)=x^2+5+2x^2-6x=3x^2-6x+5.\))

(b) \((x+1)^2\) is a factor of \(f(x)=x^3+ax^2+bx+3\).

(i) Values of a and b. If \((x+1)^2\) divides \(f(x)\), then \(x=-1\) is a repeated root, so both \(f(-1)=0\) and \(f'(-1)=0\).

\(f(-1)=-1+a-b+3=0\Rightarrow a-b=-2.\)

\(f'(x)=3x^2+2ax+b,\) so \(f'(-1)=3-2a+b=0\Rightarrow 2a-b=3.\)

Subtracting the first from the second: \(a=5.\) Then \(b=a+2=7.\)

\[a=5,\quad b=7.\]

(ii) Zeros of f(x). Now \(f(x)=x^3+5x^2+7x+3.\) Since \((x+1)^2=x^2+2x+1\) is a factor, divide:

\[x^3+5x^2+7x+3=(x+1)^2(x+3).\]

Therefore the zeros are \(x=-1\) (a double/repeated root) and \(x=-3.\)