Let our polynomial be It is clear that the coefficient of is + 2 3 + + 14 15 = P(x). x –1 – – so = 1 8x + + where is some polynomial divisible by P(x) Cx2 Q(x), Q(x) x3. – Then, = 1 + 8x + + = 1 + (2C 64)x2 + where is P(–x) Cx2 Q(–x), P(x)P(–x) R(x), R(x) – some polynomial divisible by However, we also know that x3. = (1 + + 2x)(1 2x)(1 3x)(1 + 3x) (1 15x)(1 + 15x) P(x)P(–x) x)(1 x)(1 – – – – = (1 4x2) (1 225x2) = 1 (1 + 4 + + 225)x2 + Therefore, x2)(1 R(x). – – – – 2C 64 = + 4 + + 225) = so 2C = = C – –(1 –1240, –1176, –588.
Let our polynomial be P(x). It is clear that the coefficient of x is –1 + 2 – 3 + + 14 – 15 = so P(x) = 1 – 8x + Cx2 + Q(x), where Q(x) is some polynomial divisible by x3. Then, P(–x) = 1 + 8x + Cx2 + Q(–x), P(x)P(–x) = 1 + (2C – 64)x2 + R(x), where R(x) is some polynomial divisible by x3. However, we also know that P(x)P(–x) = (1 – x)(1 + x)(1 + 2x)(1 – 2x)(1 – 3x)(1 + 3x) (1 – 15x)(1 + 15x) = (1 – x2)(1 – 4x2) (1 – 225x2) = 1 – (1 + 4 + + 225)x2 + R(x). Therefore, 2C – 64 = –(1 + 4 + + 225) = –1240, so 2C = –1176, C = –588.
Practice_3-11-12_Solutions.docx