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Worksheet: Factor Theorem and Synthetic Division

Q1:

Liam used synthetic division to prove that 4 is a root of the polynomial 𝑓 ( π‘₯ ) = 2 π‘₯ βˆ’ 9 π‘₯ + π‘₯ + 1 2 3 2 .

Using his result, factor 𝑓 ( π‘₯ ) into three linear factors.

  • A 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 4 ) ( π‘₯ βˆ’ 1 ) ( 2 π‘₯ + 3 )
  • B 𝑓 ( π‘₯ ) = ( π‘₯ + 4 ) ( π‘₯ + 1 ) ( 2 π‘₯ βˆ’ 3 )
  • C 𝑓 ( π‘₯ ) = ( π‘₯ + 4 ) ( 2 π‘₯ βˆ’ 1 ) ( π‘₯ + 3 )
  • D 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 4 ) ( π‘₯ + 1 ) ( 2 π‘₯ βˆ’ 3 )
  • E 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 4 ) ( 2 π‘₯ + 1 ) ( π‘₯ βˆ’ 3 )

Q2:

Consider the function 𝑓 ( π‘₯ ) = 2 π‘₯ βˆ’ π‘₯ βˆ’ 1 2 π‘₯ βˆ’ 7 π‘₯ βˆ’ 6 4 3 2 .

Given that two of the three numbers 1 , βˆ’ 2 , and 3 are roots of 𝑓 ( π‘₯ ) , use synthetic division to fully factor 𝑓 ( π‘₯ ) .

  • A 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 3 ) ( π‘₯ + 2 ) ( π‘₯ + 1 ) 2
  • B 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 3 ) ( π‘₯ βˆ’ 1 ) ο€Ή 2 π‘₯ + 7 π‘₯ + 1 0  2
  • C 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 3 ) ( π‘₯ + 2 ) ο€Ή π‘₯ + π‘₯ + 1  2
  • D 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 3 ) ( π‘₯ + 2 ) ο€Ή 2 π‘₯ + π‘₯ + 1  2
  • E 𝑓 ( π‘₯ ) = ( π‘₯ βˆ’ 1 ) ( π‘₯ + 2 ) ( 2 π‘₯ βˆ’ 5 ) ( π‘₯ + 1 )

Q3:

Consider the function 𝑓 ( π‘₯ ) = 2 π‘₯ + 1 0 π‘₯ + 5 π‘₯ βˆ’ 2 0 π‘₯ + 3 4 3 2 .

Using synthetic division, find the value of 𝑓 ( βˆ’ 3 ) .

State which, if any, of ( π‘₯ βˆ’ 3 ) and ( π‘₯ + 3 ) is a factor of 𝑓 ( π‘₯ ) .

  • ANeither ( π‘₯ βˆ’ 3 ) nor ( π‘₯ + 3 ) is a factor.
  • BBoth ( π‘₯ βˆ’ 3 ) and ( π‘₯ + 3 ) are factors.
  • COnly ( π‘₯ βˆ’ 3 ) is a factor.
  • DOnly ( π‘₯ + 3 ) is a factor.