Question

Please help
find all of the zeros of the function
f(x)=x^3+7x^2+4x+28

Answer 1

yes?

Answer 2

We can write as:\[x ^{3}+7x ^{2}+4x+28=x^{3}+4x+7x ^{2}+28=x(x ^{2}+4)+7(x ^{4}+4)=(x+7)(x ^{2}+4)\]

Answer 3

f(x) = 0
so
x^3+7x^2+4x+28 = 0
since 28 = 27 + 1 then
x^3 + 7x^2 + 4x + 27 + 1 = 0
and then you would try factorization of x^3 + 27
and 7x^2+4x+1
hope i helped

Answer 4

One of your best approaches would include focusing on that last term, the constant 28, and identifying its integer factors.

Are you familiar with synthetic division? That's the fastest way of determining whether a possible root of a polynomial actually is a root.

Supposing that we decide to check whether or not -4 is a root of f(x)=x^3+7x^2+4x+28.

Here's how I'd set up and execute synthetic division:

-4 | 1 7 4 28
-4 -12 32
------------
1 3 -8 60 60 is called "the remainder." If the remainder is not zero, the
possible root in question is not actually a result. Thus, -4
is not a root of f(x)=x^3+7x^2+4x+28.

If you're not familiar with synthetic division, but would be interested in learning how to use this powerful tool, please just say so.

Answer 5

my teacher used synthetic division to solve these so after you find out -4 isn't a root what do you do?

Answer 6

NooriK: "since 28 = 27 + 1" leads to the question of how this would help. Mind explaining this technique?

Answer 7

To find the zeroes:
(x+7)(x^2+4)=0
Thus x=-7 OR x=2i OR x=-2i are the Zeroes

Answer 8

Why don't we continue checking whether other factors of 28 are roots, and then, after we've found one, discuss what to do next?

Answer 9

oh you can just factor it?

Answer 10

a factor of 28 are 4 and 7

Answer 11

Certainly you can just factor the polynomial. In cases where the factoring is not so obvious, synth. div. is still probably the fastest way of identifying roots.

Let's borrow one of Rajesh's roots: x=-7.

Then the synth. div. becomes

-7 | 1 7 4 28
-7 0 -28
_____________________________
1 0 4 0

Answer 12

Since the last column (28-28) combines to produce 0, we conclude that -7 is a root.

Look at the 3 coefficients left: 1 0 4.

This represents 1x^2 + 0x +4, whose roots are plus or minus 2i, as Rajesh has concluded.

Answer 13

oh thank you i get it now

Answer 14

Should you want to use synth. div. to finish this problem, that's relatively straightforward:

2i | 1 0 4
2i -4
--------------
1 2i 0 So we know that 2i is a root (as is -2i)

Answer 15

x^3+27 = (x+3) (x^2-3x+9)
and
\[7x^2 + 4x + 1 = \frac{ -4 +\sqrt{16-28} }{ 14 }\]
or
\[7x^2 + 4x + 1 = \frac{ -4 -\sqrt{16-28} }{ 14 }\]

Answer 16

NooriK: How many roots would you expect x^3 + 7x^2 +4x + 28 to have? You seem to be implying that there are just two.