Find numbers a, b, p and q so that: x 3 + 15 x 2 + 3 x + 5 = p ( x − a ) 3 + q ( x − b ) 3

Question

Find numbers   a,   b,   p and   q so that:      x    3    +  15      x    2    +  3  x  +  5  =  p  (  x  −  a      )    3    +  q  (  x  −  b      )    3

Haggar72 · Accepted Answer

3  p      a    2    +  3  q      b    2    =  3  ⇒  p      a    2    +  q      b    2    =  1You have also   p  +  q  =  1. So   p      a    2    +  q      b    2    =  p  +  q  =  1 from here we conclude that   a  =  ±  1  ,  b  =  ±  1.  −  3  p  a  −  3  q  b  =  15  ⇒  p  a  +  q  b  =  −  5Using   p  a  +  q  b  =  −  5 and   p  +  q  =  1 with   a  =  1  ,  b  =  −  1 you find   p  =  −  2 and   q  =  +  3Using   p  a  +  q  b  =  −  5 and   p  +  q  =  1 with   a  =  −  1  ,  b  =  1 you find   p  =  3 and   q  =  −  2. Therefore the solution is  (  a  ,  b  ,  p  ,  q  )  =  (  1  ,  −  1  ,  −  2  ,  3  )    &amp;amp;    (  a  ,  b  ,  p  ,  q  )  =  (  −  1  ,  1  ,  3  ,  −  2  )

Peyton Velez · Accepted Answer

(  p  ,  q  ) and   (  p      a    2    ,  q      b    2    ) both obey the same pair of linear equations.So either the pair of equations is degenerate, or the two vectors are equal.

Find numbers a , b , p and q so that: x 3 </msup> + 1

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