quadratic equation

[Guest]

How to convert

[tex]x^2 -5x + 4 = 0[/tex]

into

(x-4)(x-1)

What is the formula?

I want also the formula to convert

[tex]x^3 + 1[/tex]

into

[tex](x+1)(x^2 - x + 1 )[/tex]

and

[tex]x^4 - 5x^2 + 4[/tex]

into

[tex](x^2 - 4 )(x^2- 1 )[/tex]

[Guest]

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

[tex]x^{2}[/tex]-5х+4=

=([tex]x^{2}[/tex]-х)+(-4х+4)=

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

=(х-1)(х-4)=

=(х-4)(х-1)

2 task
[tex]x^{3}[/tex]+1=

=[tex]x^{3}[/tex]-[tex]x^{2}[/tex]+[tex]x^{2}[/tex]+х-х+1=

=([tex]x^{3}[/tex]-[tex]x^{2}[/tex]+х)+[tex]x^{2}[/tex]-х+1=

=х([tex]x^{2}[/tex]-х+1)+([tex]x^{2}[/tex]-х+1)=

=(х+1)([tex]x^{2}[/tex]-х+1)

3 task
[tex]x^{4}[/tex]-5[tex]x^{2}[/tex]+4=

=[tex]x^{4}[/tex]-4[tex]x^{2}[/tex]-[tex]x^{2}[/tex]+4=

=[tex]x^{2}[/tex]([tex]x^{2}[/tex]-4)-1([tex]x^{2}[/tex]-4)=

=([tex]x^{2}[/tex]-4)([tex]x^{2}[/tex]-1)=(х-2)(х+2)(х-1)(х+1)

[Guest]

There is the trick for solving this question

[tex]x^{2}-5x+4=0[/tex] ( its simple)

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

In this problem it is already in the forn

After that multiply the first and the last term which is

[tex]4\times x^{2}=4x^{2}[/tex]
Now, take the combination of any two numbers which gives [tex]4x^{2}[/tex] ( 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 [tex]4x^{2}[/tex] but when we add these combination we get [tex]5x \ne -5x[/tex]) 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 [tex]4x \ne -5x[/tex]) which is not the correct combination for this ques

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

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

-x, -4x( which after multiplied gives [tex]4x^{2}[/tex] 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

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

Here is the answer

[Guest]

[tex]x^{2 }-5x+4=0[/tex]
[tex]x_{1}=1[/tex]
[tex]x_{2}=4[/tex]
[tex](x- x_{1 })(x-x_{2})=(x-1)(x-4)=x^{2}-5x+4[/tex]
[tex]x^{3}+1=0[/tex]
[tex]x_{1}=-1[/tex]
[tex]x_{2}=\frac{1}{2}+ \frac{ \sqrt{3} }{2}i[/tex]
[tex]x_{3}=\frac{1}{2}- \frac{ \sqrt{3} }{2}i[/tex]
[tex](x- x_{1 })(x-x_{2})(x-x_{3})=(x+1)(x-\frac{1}{2}- \frac{ \sqrt{3} }{2}i)(x-\frac{1}{2}+ \frac{ \sqrt{3} }{2}i)=(x+1)(x^{2}-x+1)=x^{3}+1[/tex]
[tex]x^{4}-5x^{2}+4=0[/tex]
[tex]x_{1}=1[/tex]
[tex]x_{2}=-1[/tex]
[tex]x_{3}=2[/tex]
[tex]x_{4}=-2[/tex]
[tex](x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=(x-1)(x+1)(x-2)(x+2)=(x^{2}-1)(x^{2}-4)=x^{4}-5x^{2}+4[/tex]

[Thongtm]

wow, it really great...