x？ +(a+b)x+ab

Then this equation can be decomposed into

The form of (x+a)(x-b) (a, b is a constant)

Expand:

If x? If the term has a coefficient, then there is: If the coefficient of the equation satisfies

cdx^2+(ad+cb)x+ab

It can be broken down into

(cx+a)(dx+b)

If satisfied

cdx^2+(ac+db)x+ab

Then there is

(dx+a)(cx+b)

This decomposition method starts with the constant term and decomposes the constant term into the product of two numbers.

Then decompose the coefficient of the quadratic term, and multiply the separated numbers one by one, and the sum is the coefficient of the middle term.

This equation usually gives:

x？ +(a+b)x+ab=0

now

x 1=-a

x2=-b

while

When cdx 2-(AC+DB) x+AB = 0

x 1=-b/c

X2 =- ADC

Generally speaking, any rooted equation can be decomposed into the following forms.

It's just that those roots are irrational and can't be decomposed like this

extra

Let me give you some examples:

x？ +2x-3=x？ +(3- 1)x+(-3× 1)=(x+3)(x- 1)

x？ +4x-5=x？ +(5- 1)x+(- 1×5)=(x- 1)(x+5)

x？ +7x+6=x？ +(6+ 1)x+ 1×6 =(x+6)(x+ 1)

x？ -2x+ 15=x？ +(-5+3)x+(-5×3)=(x-5)(x+3)

x？ -2x-8=x？ +(2-4)x+(-4×2)=(x-4)(x+2)

x？ - 13x+ 12=x？ +(- 1- 12)+(- 1×- 12)=(x- 1)(x- 12)

If there is anything you don't understand, or the topic won't, please ask.

In addition, there is the common factor method.

For example:

Answer? +3a=0

Then a(a+3)=0.

a=0

Or a=-3.