What are the zeros for this problem? x^4-8x^3+7x-14.

Question

This is using synthetic divsion.

Answer 1

The given expression has no rational zeroes, which would be required for synthetic division.

I suggest that the expression should have been x⁴-8x³+7x²+2x-14. When x=-1 this evaluates to:

1+8+7-2-14=0 so x=-1 is a zero. Now we can use synthetic division to reduce to a cubic:

-1 | 1 -8   7    2  -14

      1 -1   9 -16 | 14

      1 -9 16 -14 |   0 = x³-9x²+16x-14.

14=1×14=2×7, so ±7 or ±2 may be zeroes. So try ±2:

±8-36±32-14 which gives us either -10 or -90. So this is not a zero.

Now try ±7: ±343-441±112-14 which gives us either 0 or -910, therefore 7 is a zero.

Use synthetic division again:

7 | 1 -9  16  -14

     1  7 -14 | 14

     1 -2    2 |   0 = x²-2x+2 and we can find the complex zeroes:

x²-2x=-2, x²-2x+1=-1, (x-1)²=-1, x=1±i where i=√-1.

So the zeroes are -1, 7, 1+i, 1-i.