Question

Help with math question?

Answer 1

omgg

Answer 2

sorry no help

Answer 3

hmm looks like we should use p/q method

Answer 4

\[\large\rm \frac{ p }{ q }=\frac{ constant }{ leading ~coefficient } \]
constant is what in this equation ? and leading coefficient is what ?

Answer 5

uum yaa

Answer 6

1/5 ?

Answer 7

well we write with plus minus sign \[\huge\rm \frac{ \pm 1 }{ \pm 5 }\]
btw we need factor of constant term and factor of leading coefficient
factors of 1 is just 1
and factors of 5 is just 1 and 5 \[\frac{ \pm 1 }{ \pm1 , \pm 5}\]
now we should use the synthetic division to find which one is correct zero

Answer 8

pm 1/ pm 1 is one
\[\frac{ \pm 1}{ \pm 1, \pm 5 } =\frac{ \pm1 }{ \pm 1 } , \frac{ \pm 1 }{ \pm 5} \]
=\[\pm 1 , \frac{\pm 1}{\pm 5}\]

Answer 9

you can use synthetic division `or` substitute x for each number if you get 0 as a result
then the number u sub would be the zero

Answer 10

So the factor will be f(x)=2(x+1/5)(x-1)?

Answer 11

we are looking for zeros right ?

Answer 12

hmm hmmm
it's not factorable that's why we were using p/q method hmm

Answer 13

I am so confused. How I solve with synthetic division to find the zeros

Answer 14

well forget the synthetic division
sub x for each zero if you get 0 then the number u substituted would be the zero

Answer 15

\[\pm 1 , \frac{\pm 1}{\pm 5} \]\[-1, 1 , \frac{1}{5} , -\frac{1}{5}\]
sub x for each number

Answer 16

lets try first one \[0=5(1)^3+6(1)^2+6(1)+1\]zeros is a point where the graph intersect the x-axis when y = 0
so that's why i replaced f(x) which is same as y with 0
both sides are equal ??
or don't sub y for 0\[\rm f(x)=y=5(1)^3+6(1)^2+6(1)+1\]
solve right side do you get zero if yes then 1 is one of the zero

Answer 17

make sense ?

Answer 18

Yes, So the 1 is not one of the zero

Answer 19

So the only answer is -1/5

Answer 20

yes right :=))