Let P(x) = x3 + ax2 + bx + c be a polynomial where a, b, c are integers and c is odd. Let \pi be the

Question

Let P(x) = x3 + ax2 + bx + c be a polynomial where a, b, c are integers and c is odd. Let \pi be the value of P(x) at x = i. Given that p13 + p23 + p33 = 3 p1p2 p3 , find the value of p2 + 2p1 – 3p0.

Accepted Answer

{"content":[{"content":[{"marks":[{"type":"bold"}],"text":"Step 1:","type":"text"},{"text":"      p13 + p23 + p33 = 3 p1p2 p3","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 2:","type":"text"},{"text":" =>    either p1 + p2 + p3 = 0 or p1 = p2 = p3","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 3:","type":"text"},{"text":" Here p1 = ","type":"text"},{"attrs":{"latex":"a"},"type":"math_inline"},{"text":" + ","type":"text"},{"attrs":{"latex":"b"},"type":"math_inline"},{"text":" + ","type":"text"},{"attrs":{"latex":"c"},"type":"math_inline"},{"text":" + 1                                           …(i)","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 4:","type":"text"},{"text":" p2 = 4a + 2b + ","type":"text"},{"attrs":{"latex":"c"},"type":"math_inline"},{"text":" + 8                                         …(ii)","type":"text"}],"type":"paragraph"},{"type":"paragraph","content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step 5:"},{"type":"text","text":" p3 = 9a + 3b + "},{"attrs":{"latex":"c"},"type":"math_inline"},{"text":" + 27                                        …(iii)","type":"text"}]},{"content":[{"marks":[{"type":"bold"}],"text":"Step 6:","type":"text"},{"text":" Here p1 + p2 + p3 = 14a + 6b + 3c + 36","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 7:","type":"text"},{"text":" Here (14a + 6b + 36) + 3c = 0","type":"text"}],"type":"paragraph"},{"type":"paragraph","content":[{"type":"text","marks":[{"type":"bold"}],"text":"Step 8:"},{"text":" Even + odd \neq even (c is odd)","type":"text"}]},{"content":[{"marks":[{"type":"bold"}],"text":"Step 9:","type":"text"},{"text":" =>    p1 + p2 + p3 \neq 0","type":"text"}],"type":"paragraph"},{"type":"paragraph","content":[{"text":"Step 10:","type":"text","marks":[{"type":"bold"}]},{"text":" Hence, p1 = p2 = p3                                             …(iv)","type":"text"}]},{"content":[{"marks":[{"type":"bold"}],"text":"Step 11:","type":"text"},{"text":" Hence solving equation (i), (ii), (iii) and (iv) we get","type":"text"}],"type":"paragraph"},{"content":[{"marks":[{"type":"bold"}],"text":"Step 12:","type":"text"},{"text":" a = –6, ","type":"text"},{"attrs":{"latex":"b"},"type":"math_inline"},{"type":"text","text":" = 11"}],"type":"paragraph"},{"type":"paragraph","content":[{"marks":[{"type":"bold"}],"text":"Step 13:","type":"text"},{"text":" but p2 + 2p1 – 3p0 = 6a + 4b + 10 = 18","type":"text"}]}],"type":"doc"}

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