# Thread: Delta and factorization of polynomials?

1. ## Delta and factorization of polynomials?

I am wondering if Delta (b square minus 4ac) has any relation or can determine if a quadratic polynomial (with a degree of 2) can be factorized?

To put it simply, let say, a quadratic expression 30x^2 + 37x + 84 , it can either be factorized into (x-a)(x-b) where a and b are nice integers by cross method or cannot be factorized into integral a and b. So besides trial and error test with cross method, is there any quick way or using delta can determine whether it can be factorized into integral a and b in form of (x-a)(x-b)?

thanks all

2. Originally Posted by kenny1999
I am wondering if Delta (b square minus 4ac) has any relation or can determine if a quadratic polynomial (with a degree of 2) can be factorized?

To put it simply, let say, a quadratic expression 30x^2 + 37x + 84 , it can either be factorized into (x-a)(x-b) where a and b are nice integers by cross method or cannot be factorized into integral a and b. So besides trial and error test with cross method, is there any quick way or using delta can determine whether it can be factorized into integral a and b in form of (x-a)(x-b)?

thanks all
Recall that if a Quadratic can be factorised, then it will have solutions.

Therefore, if there are not any solutions, it can not be factorised.

i.e. If $\Delta < 0$ the function can not be factorised.

Also, recall that you only get one solution if the factorised form of the Quadratic is a Perfect Square.

Therefore if $\Delta = 0$ the function will be factorised into a perfect square.

3. If $\Delta\geq 0$ then you can find the roots:

$x_1=\frac{-b-\sqrt{\Delta}}{2a}, \ x_2=\frac{-b+\sqrt{\Delta}}{2a}$

Then you can factorize the quadratic using the roots in the following identity:

$ax^2+bx+c=a(x-x_1)(x-x_2)$

(The identity holds even the roots are complex numbers.)

4. Originally Posted by red_dog
If $\Delta\geq 0$ then you can find the roots:

$x_1=\frac{-b-\sqrt{\Delta}}{2a}, \ x_2=\frac{-b+\sqrt{\Delta}}{2a}$

Then you can factorize the quadratic using the roots in the following identity:

$ax^2+bx+c=a(x-x_1)(x-x_2)$

(The identity holds even the roots are complex numbers.)
True. I was assuming that the OP was factorising over the Real Number field.

5. yes, thank all for the answer

but I want to know.....

Usually by using the general cross method to factorize a quadratic polynomials x^2+bx+c, we will either be able to get (x-p)(x-q) where
p and q are integral values, or unable to factorize. But whether it can be factorized is highly determined by trial-and-error in cross method.

My question is (was) , is there any other way, rather than trial and error in the cross-method to determine whether
ax^2 + bx + c can be factorized into the form of (x-p)(x-q), where p and q are INTEGERS

I understand delta, if delta is less than 0, then the surd will be undefined in real region, so there is no value of x

and I also understand that if delta = 0 or bigger than 0 there will be real value of x, but my question is how to know if the roots of a
quadratic equation is an integer or not (not integers). Thanks all in advance

6. Originally Posted by kenny1999
yes, thank all for the answer

but I want to know.....

Usually by using the general cross method to factorize a quadratic polynomials x^2+bx+c, we will either be able to get (x-p)(x-q) where
p and q are integral values, or unable to factorize. But whether it can be factorized is highly determined by trial-and-error in cross method.

My question is (was) , is there any other way, rather than trial and error in the cross-method to determine whether
ax^2 + bx + c can be factorized into the form of (x-p)(x-q), where p and q are INTEGERS

I understand delta, if delta is less than 0, then the surd will be undefined in real region, so there is no value of x

and I also understand that if delta = 0 or bigger than 0 there will be real value of x, but my question is how to know if the roots of a
quadratic equation is an integer or not (not integers). Thanks all in advance
If $\Delta$ is a perfect square, then you will be able to factorise so that $p$ and $q$ are integers.

7. Assume ax^2 + bx + c = a(x-p)(x-q)
= ax^2 -a(p+q) x + apq

Equate the coefficients
p + q = -b/a
pq = c/a
check if b/a is integer or not and c/a is integer or not..

if they are not.. then p and q are not integer..

if they are integer..see if you can solve them for integers...

probably this is what you were looking for.

8. Originally Posted by Prove It
If $\Delta$ is a perfect square, then you will be able to factorise so that $p$ and $q$ are integers.
why you can conclude that?

delta is only PART of the story

it has denominator 2a

an integral value divides an integral value 2a not necessarily an integers.

This is my question.

9. Originally Posted by kenny1999
why you can conclude that?

delta is only PART of the story

it has denominator 2a

an integral value divides an integral value 2a not necessarily an integers.

This is my question.
I should have said rational numbers.

Having said that...

If $\Delta$ is a perfect square, then you have rational roots of the form $x = \frac{a}{b}$.

Therefore $bx - a$ is a factor.

So you can still have an integer value of your $p$ and $q$. You will also have an integer multiple of $x$.

10. Originally Posted by tutor.mathitutor
Assume ax^2 + bx + c = a(x-p)(x-q)
= ax^2 -a(p+q) x + apq

Equate the coefficients
p + q = -b/a
pq = c/a
check if b/a is integer or not and c/a is integer or not..

if they are not.. then p and q are not integer..

if they are integer..see if you can solve them for integers...

probably this is what you were looking for.
i think you understand what i am asking for, thanks

but how can you conclude that if p+q is an integer
and p x q is an integer, then p and q itself must also be an integer?

11. Originally Posted by Prove It
I should have said rational numbers.

Having said that...

If $\Delta$ is a perfect square, then you have rational roots of the form $x = \frac{a}{b}$.

Therefore $bx - a$ is a factor.

So you can still have an integer value of your [tex]p[\math] and $q$. You will also have an integer multiple of $x$.
can i conclude that

if delta < 0

the quadratic expression cannot be factorised

while if delta = 0

it can be factorised into (x-p)^2 where p is rational

and while if delta > 0 and is a perfect square

then it can be factorised into (x-p)(x-q) where p and q are rational

and while if delta > 0 and not a perfect square

then it can also be factorised into (x-p)(x-q) but p and q are irrational

??

12. Originally Posted by kenny1999
can i conclude that

if delta < 0

the quadratic expression cannot be factorised

while if delta = 0

it can be factorised into (x-p)^2 where p is rational

and while if delta > 0 and is a perfect square

then it can be factorised into (x-p)(x-q) where p and q are rational

and while if delta > 0 and not a perfect square

then it can also be factorised into (x-p)(x-q) but p and q are irrational

??
It would actually be

If $\Delta > 0$ and a perfect square, then it can be factorised into $(mx - p)(nx - q)$ where $m, n, p, q$ are rational numbers.

If $\Delta > 0$ and not a perfect square, then it can be factorised into $(mx - p)(nx - q)$ where $m, n, p, q$ are irrational numbers.

13. Originally Posted by Prove It
I should have said rational numbers.

Having said that...

If $\Delta$ is a perfect square, then you have rational roots of the form $x = \frac{a}{b}$.

Therefore $bx - a$ is a factor.

So you can still have an integer value of your $p$ and $q$. You will also have an integer multiple of $x$.
you are right, but I have a problem

but not quadratic eqauation i.e. ax^2+bx+c=0

equation can multiply the denominator to both sides
but expression cannot.

14. Originally Posted by Prove It
It would actually be

If $\Delta > 0$ and a perfect square, then it can be factorised into $(mx - p)(nx - q)$ where $m, n, p, q$ are rational numbers.

If $\Delta > 0$ and not a perfect square, then it can be factorised into $(mx - p)(nx - q)$ where $m, n, p, q$ are irrational numbers.
ok
but why?
can you prove to me? thanks!

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