Grades 11-12 Video Solutions 2024-2024_11-12

Video Transcription

Polynomial p of x satisfies the relationship p of x plus 1 equals x squared minus x plus 2p of 6 for every real x. What is the sum of the coefficients of p?  First note that substituting x equals 5 in gives us p6 equals 5 squared minus 5 plus 2p6, and then solving for p6, we can then obtain p6 equals negative 20.  Putting this back into the relation gives us px plus 1 equals x squared minus x minus 40. Thus, the sum of coefficients of p is equal to p1, which is p1 equal to p0 plus 1 equal  to 0 squared minus 0 minus 40, which is negative 40, our answer.

Video Summary

The polynomial \( p(x) \) satisfies the condition \( p(x + 1) = x^2 - x + 2p(6) \) for all real \( x \). By substituting \( x = 5 \), we find \( p(6) = -20 \). Substituting back, we get \( p(x + 1) = x^2 - x - 40 \). The sum of the coefficients of \( p(x) \) is determined by evaluating \( p(1) \), which equals \( p(0 + 1) = 0^2 - 0 - 40 \), giving a sum of \(-40\). Thus, the sum of the coefficients of \( p \) is \(-40\).

Keywords

polynomial

coefficients

substitution

evaluation

sum