How to convert

$$x^2 -5x + 4 = 0$$

into

(x-4)(x-1)

What is the formula?

I want also the formula to convert

$$x^3 + 1$$

into

$$(x+1)(x^2 - x + 1 )$$

and

$$x^4 - 5x^2 + 4$$

into

$$(x^2 - 4 )(x^2- 1 )$$
Guest

It is veri easy
1 task ( I will solve them without formulas.)

$$x^{2}$$-5х+4=

=($$x^{2}$$-х)+(-4х+4)=

=х(х-1)-4(х-1)= [ There is a common multiplier. ]

=(х-1)(х-4)=

=(х-4)(х-1)

$$x^{3}$$+1=

=$$x^{3}$$-$$x^{2}$$+$$x^{2}$$+х-х+1=

=($$x^{3}$$-$$x^{2}$$+х)+$$x^{2}$$-х+1=

=х($$x^{2}$$-х+1)+($$x^{2}$$-х+1)=

=(х+1)($$x^{2}$$-х+1)

$$x^{4}$$-5$$x^{2}$$+4=

=$$x^{4}$$-4$$x^{2}$$-$$x^{2}$$+4=

=$$x^{2}$$($$x^{2}$$-4)-1($$x^{2}$$-4)=

=($$x^{2}$$-4)($$x^{2}$$-1)=(х-2)(х+2)(х-1)(х+1)
Guest

There is the trick for solving this question

$$x^{2}-5x+4=0$$ ( its simple)

Firstly arrange these types of functions in the given order
$$ax^{2}+bx+c=0$$

In this problem it is already in the forn

After that multiply the first and the last term which is

$$4\times x^{2}=4x^{2}$$
Now, take the combination of any two numbers which gives $$4x^{2}$$ ( multiplication of first and last term ) after multiplied and givers the middle term i.e. -5x after get added

lets take the combination, for this ques

We can take the combination of :

x, 4x( which after multiplied gives $$4x^{2}$$ but when we add these combination we get $$5x \ne -5x$$) which is not the correct combination for this ques

2x, 2x(which after multiplied gives 4x^{2} but when we add these combination we get $$4x \ne -5x$$) which is not the correct combination for this ques

-2x, -2x(which after multiplied gives $$4x^{2}$$ but when we add these combination we get $$-4x \ne -5x$$) which is not the correct combination for this ques

-2x, x( which after multiplied gives $$-4x^{2} \ne 4x^{2}$$)this is also not the correct option

-x, -4x( which after multiplied gives $$4x^{2}$$ but when we add these combination we get -5 = -5x) so, this is the correct combination for this ques

Now, replace the middle term (-5x) by the correct combination (-x -4x) (which you got by the previous step )

By replacing we get

$$x^{2}-x-4x+4=0$$
$$x(x-1)-4(x-1)=0$$[ there is a common multiplier ]
$$(x-1)(x-4)=0$$
$$\Rightarrow$$
$$(x-4)(x-1)=0$$