1 problem found
Prove that if \({(x-a)^{2}}\) is a factor of the polynomial \(\p(x)\), then \(\p'(a)=0\). Prove a corresponding result if \((x-a)^4\) is a factor of \(\p(x).\) Given that the polynomial $$ x^6+4x^5-5x^4-40x^3-40x^2+32x+k $$ has a factor of the form \({(x-a)}^4\), find \(k\).
Solution: First notice that \(p(x) = (x-a)^2q(x)\) so \(p'(x) = 2(x-a)q(x) + (x-a)^2q'(x) = (x-a)(2q(x)+(x-a)q'(x))\), in particular \(p'(a) = 0\) so \(x-a\) is a root of \(p'(x)\). If \((x-a)^4\) is a root of \(p(x)\) then \(p^{(3)}(a)= 0\). The proof is similar. Differentiating \(3\) times we obtain: \(6 \cdot 5 \cdot 4 x^3 + 4 \cdot 5 \cdot 4 \cdot 3 x^2 - 5\cdot4 \cdot 3 \cdot 2 x-40 \cdot 3 \cdot 2 \cdot 1 = 5!(x^3+2x^2-x-2) = 5!(x+2)(x^2-1)\). So our possible (repeated) roots are \(x=-2,-1,1\). We can check \(p'(x) = 6x^5+20x^4-20x^3-120x^2-80x+32\), and see \(p'(1) = 36 - 200 \neq 0\), \(p'(-1) = -6+20+20-120+80+32 \neq 0\), therefore \(a = -2\)