Prove that, if \(\alpha\) is a repeated root of \(f(x)=0\), where \(f(x)\) is a polynomial, then \(\alpha\) is a root of the equation \(f'(x)=0\).

Since we know that \(f(x)=0\) has a repeated root \(\alpha\), we may write \[f(x)=(x-\alpha)^2 g(x),\] where \(g(x)\) is some polynomial. By the product rule we have \[f'(x)=2(x-\alpha)g(x)+(x-\alpha)^2 g'(x)=(x-\alpha)\left[2g(x)+(x-\alpha)g'(x)\right],\] so we have that \(\alpha\) is a root of \(f'(x)=0\).

Given that the equation \(4x^4+x^2+3x+1=0\) has a repeated root, find its value.

If \(f(x)=4x^4+x^2+3x+1\), then \[f'(x)=16x^3+2x+3.\]

We are told that \(f(x) = 0\) has a repeated root, which we may call \(\alpha\). From the above, we now know that \(\alpha\) is a root of \(f'(x) = 0\).

So we seek a root of \(16x^3+2x+3=0\).

Clearly the root \(x\) can’t be a whole number, since \(16x^3+2x\) would be even and \(3\) odd.

The coefficients \(16\) and \(2\) make us think the substitution \(x = \dfrac{y}{2}\) would be helpful, where \(y\) is a whole number.

That gives \(2y^3 + y + 3 = 0\), and clearly \(y = -1\) is a root here. We have \(2y^3 + y + 3 = (y + 1)(2y^2 -2y + 3)\).

Note the roots of the quadratic bracket are complex, since \(b^2 < 4ac\).

So the repeated root we seek must be \(-\dfrac{1}{2}\). We can check this, finding

\[f(x) = (2x + 1)(2x^3 - x^2 + x + 1) = (2x + 1)^2(x^2 - x + 1).\]

Alternatively, we could note that \(f'(0)\) = 3, and \(f'(-1) = -15\), so \(f'(x) = 0\) must have a root between \(-1\) and \(0\), since \(f'(x)\) is continuous and we have a change of sign.

The value \(-1/2\) is a sensible choice to test as a possible root, and it works.