Question

how to factor a polynomial to find zeros and multiplicity. polynomial is x^4-7x^3+12x^2+4x-16

Answer 1

try factors of -16 to see if you get remainder 0 or you could plug into the function and see if the output is 0 if the output is 0 then that number that you plugged in is a zero

Answer 2

use the factor theorem
eg try f(1) - if this is 0 then x-1 is a factor:
f(1) = 1 - 7 + 12 + 4 -16 = -6 - no))
f(-1) = 1 + 7 + 12 - 4 - 16 = 0 - so (x+ 1) is a factor

now divide the polynomial by x+ 1
to give a cubic expression and comtinue as above

Answer 3

lets try -4
-4 | 1 -7 12 4 -16
| -4 -44
-------------------
1 -11 -32
so this is defintely not gonna give us remainder 0

lets try x=-2

-2| 1 -7 12 4 -16
| -2 18
-----------------
1 -9 30
this will not work either

lets try x=1

1 | 1 -7 12 4 -16
| 1 -6 6 10
--------------------
1 -6 6 10| -6
this will not work either since remainder is -6

try x=-1

-1 | 1 -7 12 4 -16\
| -1 8 -20 16
--------------------------
1 -8 20 -16 | 0
so x=-1 works which means x+1 is a factor
and x^3-8x^2+20x-16 is also a factor

Answer 4

to give a trinomial which you may be able to factor

Answer 5

lets try x=2

2 | 1 -8 20 -16
| 2 -12 16
---- -------------------------
1 -6 8 | 0
so x=2 works which implies x-2 is also a factor
and x^2-6x+8 is another factor

Answer 6

you should be able to get the rest

Answer 7

since we have
x^4-7x^3+12x^2+4x-16=(x+1)(x-2)(x^2-6x+8)

Answer 8

so are the factors of x^2-6x+8

Answer 9

what*