Algebra II 5.

Question

Find the zeros of the polynomial and state the multiplicity of each zero.
f(x) = x4 - 25x2 + 144
Question 5 options:

1) x = 16, multiplicity 1; x = 9, multiplicity 1

2) x = 16, multiplicity 2; x = 3, multiplicity 1

3) x = -4, multiplicity 1; x = 4, multiplicity 1; x = -3, multiplicity 1; x = 3, multiplicity 1

4) x = 16, multiplicity 2; x = 9, multiplicity 2

Answer 1

Hi Carrie,

 

Algebra II 5.
Find the zeros of the polynomial and state the multiplicity of each zero.
f(x) = x⁴ - 25x² + 144
Question 5 options:

1) x = 16, multiplicity 1; x = 9, multiplicity 1

2) x = 16, multiplicity 2; x = 3, multiplicity 1

3) x = -4, multiplicity 1; x = 4, multiplicity 1; x = -3, multiplicity 1; x = 3, multiplicity 1

4) x = 16, multiplicity 2; x = 9, multiplicity 2

 

 

f(x) = x⁴ - 25x² + 144

 

Need to factor the polynomial first,

Since the 3rd term is positive, the both factors must have the same sign.

Since the 2nd term is negative, then both factors must be negative.

 

Now you need to find factors of 144 which add up to -25

 

-1   -144   no

-2    -72   no

-3     -48   no

-4     -36    no   -4 + -36 =-40 but getting closer

-6     -24    no

-8     -18   no   -8+-18=-26

-9     -16   YES   -9+-16=-25

 

 

f(x) = x⁴ - 25x² + 144

 

      = (x²-9)(x²-16)

 

Both these factors can be factored.

Both are the difference of perfect squares which a²-b²=(a-b)(a+b)

 

So, x²-9=(x-3)(x+3)

and x²-16=(x-4)(x+4)

 

f(x) = x4 - 25x2 + 144

      = (x2-9)(x2-16)

      =(x-3)(x+3)(x-4)(x+4)

 

Now set each factor equal to zero and solve:

 

x-3=0

x=3

 

x+3=0

x=-3

 

x-4=0

x=4

 

x+4=0

x=-4

 

There are no repeat zeroes so the multiplicity is 1.

 

ANSWER:  x = -4, multiplicity 1; x = 4, multiplicity 1; x = -3, multiplicity 1; x = 3, multiplicity 1  (Option #3)