1.16: Factor and value condition on a cubic

0606/13/O/N/20 — Question 7 · 8 marks

The polynomial p(x)=ax3+bx2−19x+4p(x) = ax^{3} + bx^{2} - 19x + 4, where aa and bb are constants, has a factor x+4x + 4 and is such that 2p(1)=5p(0)2p(1) = 5p(0).
(a)
Show that p(x)=(x+4)(Ax2+Bx+C)p(x) = (x + 4)(Ax^{2} + Bx + C), where AA, BB and CC are integers to be found.
[6]
(b)
Hence factorise p(x)p(x).
[1]
(c)
Find the remainder when p′(x)p'(x) is divided by xx.
[1]