Worksheet: Remainder and Factor Theorem with Synthetic Division

In this worksheet, we will practice identifying factors and zeros and finding the remainder of a polynomial function using the remainder and factor theorem with synthetic division.

Q1:

Liam used synthetic division to prove that 4 is a root of the polynomial 𝑓(π‘₯)=2π‘₯βˆ’9π‘₯+π‘₯+12.

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

  • A𝑓(π‘₯)=(π‘₯+4)(π‘₯+1)(2π‘₯βˆ’3)
  • B𝑓(π‘₯)=(π‘₯βˆ’4)(π‘₯βˆ’1)(2π‘₯+3)
  • C𝑓(π‘₯)=(π‘₯βˆ’4)(2π‘₯+1)(π‘₯βˆ’3)
  • D𝑓(π‘₯)=(π‘₯βˆ’4)(π‘₯+1)(2π‘₯βˆ’3)
  • E𝑓(π‘₯)=(π‘₯+4)(2π‘₯βˆ’1)(π‘₯+3)

Q2:

Consider the function 𝑓(π‘₯)=2π‘₯+10π‘₯+5π‘₯βˆ’20π‘₯+3οŠͺ.

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‘₯).

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

Q3:

Consider the function 𝑓(π‘₯)=2π‘₯βˆ’π‘₯βˆ’12π‘₯βˆ’7π‘₯βˆ’6οŠͺ.

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‘₯).

  • A𝑓(π‘₯)=(π‘₯βˆ’1)(π‘₯+2)(2π‘₯βˆ’5)(π‘₯+1)
  • B𝑓(π‘₯)=(π‘₯βˆ’3)(π‘₯+2)(π‘₯+1)
  • C𝑓(π‘₯)=(π‘₯βˆ’3)(π‘₯+2)ο€Ήπ‘₯+π‘₯+1ο…οŠ¨
  • D𝑓(π‘₯)=(π‘₯βˆ’3)(π‘₯+2)ο€Ή2π‘₯+π‘₯+1ο…οŠ¨
  • E𝑓(π‘₯)=(π‘₯βˆ’3)(π‘₯βˆ’1)ο€Ή2π‘₯+7π‘₯+10ο…οŠ¨

Q4:

One of the zeros of the function 𝑓(π‘₯)=π‘₯βˆ’4π‘₯βˆ’17π‘₯+60 belongs to the set {2,3,4}. Using synthetic division, find all the zeros of 𝑓.

  • Aβˆ’4, 2, 6
  • Bβˆ’4, 3, βˆ’5
  • Cβˆ’4, 3, 5
  • D4, 3, βˆ’5
  • E4, 2, βˆ’6

Q5:

The function 𝑓(π‘₯)=π‘₯βˆ’5π‘₯+7π‘₯+3π‘₯βˆ’10οŠͺ has 2 real zeros and 2 imaginary zeros.

Using synthetic substitution, determine which of the values 1, 2, 3, and 4 is a root of 𝑓(π‘₯) and write 𝑓(π‘₯) in the form (π‘₯βˆ’π‘Ž)𝑄(π‘₯).

  • A𝑓(π‘₯)=(π‘₯βˆ’4)ο€Ήπ‘₯βˆ’π‘₯+3π‘₯+15ο…οŠ©οŠ¨
  • B𝑓(π‘₯)=(π‘₯+2)ο€Ήπ‘₯βˆ’7π‘₯+21π‘₯βˆ’39ο…οŠ©οŠ¨
  • C𝑓(π‘₯)=(π‘₯βˆ’2)ο€Ήπ‘₯βˆ’3π‘₯+π‘₯+5ο…οŠ©οŠ¨
  • D𝑓(π‘₯)=(π‘₯+4)ο€Ήπ‘₯βˆ’9π‘₯+43π‘₯βˆ’175ο…οŠ©οŠ¨
  • E𝑓(π‘₯)=(π‘₯βˆ’1)ο€Ήπ‘₯βˆ’4π‘₯+3π‘₯+6ο…οŠ©οŠ¨

Using synthetic substitution, determine which of the values βˆ’1, βˆ’2, βˆ’3, and βˆ’4 is a root of 𝑄(π‘₯) and write 𝑄(π‘₯) in the form (π‘₯βˆ’π‘)𝑃(π‘₯).

  • A𝑄(π‘₯)=(π‘₯+1)ο€Ήπ‘₯βˆ’4π‘₯+5ο…οŠ¨
  • B𝑄(π‘₯)=(π‘₯βˆ’1)ο€Ήπ‘₯βˆ’2π‘₯βˆ’1ο…οŠ¨
  • C𝑄(π‘₯)=(π‘₯+1)ο€Ήπ‘₯βˆ’8π‘₯+29ο…οŠ¨
  • D𝑄(π‘₯)=(π‘₯βˆ’2)ο€Ήπ‘₯βˆ’π‘₯βˆ’1ο…οŠ¨
  • E𝑄(π‘₯)=(π‘₯+2)ο€Ήπ‘₯βˆ’5π‘₯+11ο…οŠ¨

State all the zeros of 𝑓.

  • Aβˆ’2, 1, 2+𝑖, 2βˆ’π‘–
  • B2, βˆ’1, 2+𝑖, 2βˆ’π‘–
  • C4, βˆ’1, 2+𝑖, 2βˆ’π‘–
  • D4, 1, 52+√192𝑖, 52βˆ’βˆš192𝑖
  • E2, βˆ’1, 52+√192𝑖, 52βˆ’βˆš192𝑖

Q6:

A polynomial 𝑓(π‘₯) is divided by (π‘₯βˆ’π‘Ž). Given that (π‘₯βˆ’π‘Ž) is not a factor of 𝑓(π‘₯), what is the remainder equal to?

  • A𝑓(π‘Ž)
  • B𝑓(π‘Ÿ)
  • C0
  • D𝑓(π‘₯)

Q7:

Given that (π‘₯βˆ’π‘Ž) is a factor of 𝑓(π‘₯), what is the remainder when 𝑓(π‘₯) is divided by (π‘₯βˆ’π‘Ž)?

Q8:

Consider the function 𝑓(π‘₯)=π‘₯βˆ’9π‘₯+3π‘₯βˆ’7π‘₯+12οŠͺ.

Use synthetic division to find the quotient 𝑄(π‘₯) and the remainder 𝑅 satisfying 𝑓(π‘₯)=𝑄(π‘₯)(π‘₯+2)+𝑅.

  • A𝑄(π‘₯)=π‘₯βˆ’11π‘₯+25π‘₯βˆ’57, 𝑅=102
  • B𝑄(π‘₯)=π‘₯+7π‘₯βˆ’11π‘₯+15, 𝑅=βˆ’42
  • C𝑄(π‘₯)=π‘₯βˆ’11π‘₯+25π‘₯βˆ’57, 𝑅=126
  • D𝑄(π‘₯)=π‘₯βˆ’7π‘₯βˆ’11π‘₯βˆ’29, 𝑅=βˆ’46
  • E𝑄(π‘₯)=π‘₯βˆ’7π‘₯βˆ’11π‘₯βˆ’29, 𝑅=βˆ’70

Find 𝑓(βˆ’2).

Q9:

Use synthetic substitution to find the value of 𝑓(20) given 𝑓(π‘₯)=0.06π‘₯βˆ’0.14π‘₯βˆ’3.1π‘₯+5.4οŠͺ.

Q10:

Consider the function 𝑓(π‘₯)=π‘₯+5π‘₯βˆ’8π‘₯+9οŠͺ.

What does the remainder theorem tell us about 𝑓(3)?

  • A𝑓(3) is the remainder when we divide 𝑓(π‘₯) by 3π‘₯βˆ’3.
  • B𝑓(3) is the remainder when we divide 𝑓(π‘₯) by π‘₯βˆ’3.
  • C𝑓(3) is the remainder when we divide 𝑓(π‘₯) by π‘₯.
  • D𝑓(3) is the remainder when we divide 𝑓(π‘₯) by π‘₯+3.

Hence, use synthetic division to find 𝑓(3).

Q11:

Find the value of π‘Ž given that 2π‘₯+π‘Žπ‘₯βˆ’21π‘₯βˆ’36 is divisible by (π‘₯+4).

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