Question

Simplify the expression:

\[28x^2+4x-9\]

Answer 1

hi purplec
do u think of factorise it?

Answer 2

When the coefficient in front of the x^2 is larger than 4 I tend to get stuck... I just figured it out but do you know a short-cut or a tip to go about factorizing quadratic and cubics... when the coefficient in front of the highest degree is larger.

Answer 3

Use the quadratic formula. If the discriminant is a perfect square, then that will tell you if the quadratic is factorable or not


In this case, the discriminant is


D = b^2 - 4ac

D = 4^2 - 4*28*(-9)

D = 1024

which is a perfect square (since 32^2 = 1024)


So 28x^2+4x-9 can be factored.


Once you know it can be factored, keep using the quadratic formula to find the two roots of the quadratic (say r1 and r2)


The quadratic 28x^2+4x-9 will then factor to (x - r1)(x - r2)

Answer 4

With cubics, you would use the rational root theorem and polynomial long division to factor.

Answer 5

So if the discriminant is not factorable what will D be @jim_thompson5910, I know one conditions will be imaginary is there another?

Answer 6

Or does it have to be a square?

Answer 7

If it can't be factored, then D won't be a perfect square or D will be negative.

Answer 8

Okay, but how do I find my factors?...

Answer 9

What do you get when you solve 28x^2+4x-9 = 0 for x (using the quadratic formula)?

Answer 10

1/2 and -9/14

Answer 11

so x = 1/2 or x = -9/14


Since x = 1/2, we can say...


x = 1/2

2x = 1

2x - 1 = 0


because x = -9/14, we can also say...


x = -9/14

14x = -9

14x + 9 = 0

-----------------------------------

So if x = 1/2 or x = -9/14, then


2x - 1 = 0 or 14x + 9 = 0


------------------------------------

From here, use the zero product property to go from


2x - 1 = 0 or 14x + 9 = 0

to

(2x - 1)(14x + 9) = 0


So because solving the equation above gives you x = 1/2 or x = -9/14, the same solutions as the original quadratic, this means that


28x^2+4x-9 = (2x - 1)(14x + 9)


Note: this jump is possible because the GCF of the left side is 1

Answer 12

Oh... I see okay... so how would I use the use the rational root theorem and polynomial long division to factor cubics?

Answer 13

The rational root theorem is basically a tool to pick out all the possible rational roots. In other words, it allows you to list all the possible roots, which you can check if they actually are roots or not.

Answer 14

Once you've found a rational root (if there is one), you would then use it to construct a factor. Then you would use this factor to divide and find the other factor.

Answer 15

Oh...will you show me with an example? Please :D.

Answer 16

ex:

Use the rational root theorem to find the possible rational roots of x^3-6x^2+12x-8


So list the factors of 8 (the last term): 1, 2, 4, 8, -1, -2, -4, -8

and list the factors of the first coefficient (1): 1, -1


Then divide each factor of the last term by each factor of the first term to get the possible rational roots to be...


1, 2, 4, 8, -1, -2, -4, -8


So that list above is the set of possible rational roots. If x^3-6x^2+12x-8 has a rational root, then it HAS to be one (or more) of those numbers

Answer 17

So you would then plug each rational root into x^3-6x^2+12x-8 and see which one gives you zero.

It turns out that plugging x = 2 gives you zero. So

x = 2 is a root

which means

x-2 is a factor

Answer 18

You then use polynomial long division to divide x^3-6x^2+12x-8 by x-2 to find the other factor

Answer 19

And then I should be able to do what I did earlier to find the... factors of the quadratic... cool!!!

Answer 20

________x^2 - 4x + 4 _______
x-2 | x^3- 6x^2 + 12x - 8
x^3 -2x^2
-----------
-4x^2+12x
-4x^2+8x
------------
4x - 8
4x - 8
------
0


So the other factor is x^2 - 4x + 4


Then yes, you are correct. You would use the quadratic formula to factor further. You may just get lucky and recognize/remember the factorization to save time.

Answer 21

is it (x-2)^2

Answer 22

yes, so x^3-6x^2+12x-8 completely factors to (x-2)^3

Answer 23

Oh wow... haha that's cool. Thanks so much for all your help! I really do appreciate it!

Answer 24

you're welcome