f(x)=18x^5-45x^4+28x^3+11x^2-16x+4

Question

Find all the zeros and state the multiplicity of each zero.

Answer 1

The first step is to check for "easy" zeroes, such as x=1 or x=-1.

f(1)=18-45+28+11-16+4=61-61=0 so x=1 is a zero and we can use synthetic division to reduce the polynomial:

1 | 18 -45  28 11 -16    4

     18  18 -27   1  12 | -4

     18 -27    1 12   -4 |  0 = 18x⁴-27x³+x²+12x-4. Again x=1 is a zero because: 18-27+1+12-4=0. So x=1 is a duplicate zero. Divide by it:

1 | 18 -27    1 12   -4

     18  18   -9 -8 |   4

     18   -9   -8  4 |   0 = 18x³-9x²-8x+4. This time x=1 doesn't produce a zero, neither does x=-1.

Next we look for rational zeroes by examining the factors of the coefficients of the highest and lowest degrees: 18 and 4. 18=(1,18), (2,9), (3,6) and 4=(1,4), (2,2). These are factor pairs, but we have a cubic so we need to find 3 factors: (1,18)=(1,1,18); (2,9)=(1,2,9), (2,3,3); (3,6)=(1,3,6), (1,2,9); (1,4)=(1,1,4), (1,2,2); (2,2)=(1,2,2). So, removing duplicates and listing:

18=(1,1,18), (1,2,9), (2,3,3), (1,3,6); 4=(1,1,4), (1,2,2).

We only have to find one further zero, because the polynomial will be reduced to a quadratic which can be solved by using the formula or completing the square.

If we label the highest coefficient factors a₁, a₂, a₃ and the lowest b₁, b₂, b₃, then the zeroes will be ±b_i/a_j where i and j are between 1 and 3. Here I will jump to a couple of combinations of a and b which create a true zero: b=1 and a=2, so the zero is ±½; b=2 and a=3, so the zero is ±⅔.

Let's try x=½: f(½)=18(½)³-9(½)²-8(½)+4=18/8-9/4-4+4=0. So x=½ is a zero, which we can divide by:

½ | 18 -9 -8    4

      18  9  0 | -4

      18  0 -8 |  0 = 18x²-8=2(9x²-4)=2(3x-2)(3x+2), giving zeroes ⅔ and -⅔.

Zeroes are: 1 (twice), ½, ⅔, -⅔.