Question

If (x-a) is a factor of the expression ax^3-3x^2-5ax-9,find the possible values of a.Factorise the expression for each of the values of a.

Answer 1

@Zepdrix

Answer 2

If (x-a) is a factor of the polynomial,
then x=a is a root of the polynomial.

Plugging in a should give us zero,
\(\large\rm a\cdot a^3-3a^2-5a(a)-9=0\)

Answer 3

I tried using the method u taught me before. ^

Answer 4

Does this end up being some weird thing? Hmm let's see..

Answer 5

\(\large\rm a^4-3a^2-5a^2-9=0\)

Combining like-terms,
\(\large\rm a^4-8a^2-9=0\)

Answer 6

Hmm this one works out pretty nicely I think.

-9 = -9*1

and

-8 = -9+1

Answer 7

So ya, we'd have to apply that little trick from last time,
\(\large\rm u=a^2\)

Answer 8

yep,I got
\((a^2-9)(a^2+1)\)

Answer 9

\(\large\rm a^4-8a^2-9=0\)
Ok so we've done this so far,
\(\large\rm (a^2-9)(a^2+8)=0\)
don't forget about conjugates! :)
\(\large\rm (a-3)(a+3)(a^2+8)=0\)

Answer 10

So that takes care of the second part, factorize completely.

To find values for a,
we can apply our Zero-Factor Property,
setting each individual factor equal to zero,
\(\large\rm (a-3)=0\)
\(\large\rm (a+3)=0\)
\(\large\rm (a^2+8)=0\)

Answer 11

And solve for a in each case.

Answer 12

I thought it should be
\((a^2-9)(a^2+\color{red}{1})\)

Answer 13

Ahhh sorry >.<

Answer 14

Did you understand the conjugates part though?
How we got to (a-3)(a+3) ?

Answer 15

It's okay ...yes,I understand.
I managed to factorise the expression
\((a-3)(a+3)(a+1)(a-1)\)

Answer 16

No no, sum of squares does not factor like that.
\(\large\rm a^2+1\ne(a+1)(a-1)\)

Answer 17

is it \(\(a+1)^2\) ?

Answer 18

\(\large\rm a^2+1=a^2+1\)
Can't do anything nice with it.

if you apply your zero-factor property, then you might see why.

\(\large\rm a^2+1=0\)
\(\large\rm a^2=-1\)
\(\large\rm a=\pm\sqrt{-1}\)

This gives us no real solutions right?

Answer 19

oh yesss... ^

Answer 20

So then this is our full factorization,

\(\large\rm (a-3)(a+3)(a^2+1)\)

Answer 21

my answer in the book is a little bit different. xD
answer:
a=3,\(3(x-3)(x+1)^2\),a=-3,\(-3(x+3)(x-1)^2\)

Answer 22

whu 0_o

Answer 23

Ohhh I see.
My bad.

I completely forgot about the expression involving x.

Answer 24

\(\large\rm (a-3)(a+3)(a^2+1)\)
From that factorization, we conclude that the values of a are 3 and -3.

So now we need to plug these values in for our a's,
and factorize that expression,

\(\large\rm \color{indianred}{a}x^3-3x^2-5\color{indianred}{a}x-9\)

So for a=3,

\(\large\rm \color{indianred}{3}x^3-3x^2-15x-9\)

Answer 25

okay

Answer 26

Hmm we have an ugly cubic so we can umm...

Answer 27

Oh, well we already know that (x-a) is a factor of this polynomial.
So in this case, (x-3) is a factor.

So we can do long division or synthetic.

Answer 28

This problem is pretty brutal, really long lol

Answer 29

Yep xD

Answer 30

Can I use calculator for now? xD

Answer 31

ya

Answer 32

Okay,it will be easier. ^

Answer 33

I got (x-3)(x+1)

Answer 34

Hmm,something is wrong here.
My calculator is showing two solution

Answer 35

We can start by factoring a 3 out of everything,

\(\large\rm 3(x^3-x^2-5x-3\)

And then pulling the (x-3) factor out gives us,
\(\large\rm 3(x-3)(x^2+2x+1)\)

Which factors into,
\(\large\rm 3(x-3)(x+1)^2\)

Answer 36

Then,we do the samething for a=-3

Answer 37

Ya i guess so XD

Answer 38

okay,I think I know how to do.
Thank you! @Zepdrix

Answer 39

cool :)