Respuesta :
So say for example we have
x5−3x3−2x2+6
We group the terms together by 2's
(x5−3x3)−(2x2−6)
Notice how that highlighted '-' is there...it WAS '+' but since we have a '-' in front of the parenthesis...when that distributes..it turns that '+' '-' again So now...we factor out of each parenthesis what we can... x^5 - 3x^3 ...well looks like at least x^3 can come out of there 2x^2 - 6 ...well we can factor a 2 out of there.. so now we have
x3(x2−3)−2(x2−3)
Now notice how we have 2 parenthesis that have the same thing inside them...and we also have 2 coefficients of those parenthesis...so we put those together respectively
(coefficients in here)(those things in the parentheis since they are the same)
(x3−2)(x2−3)
Everything make sense?
Well say we have
3x2+10x−8
We can actually break up that 10x ...and make it -2x + 12x right? since -2 + 12 =10 so we would then have
3x2−2x+12x−8
NOW we can group these up by 2's
(3x2−2x)+(12x−8)
Lets factor out an 'x' of the first set of parenthesis...and also lets factor out a 4 of the last set...
x(3x−2)+4(3x−2)
Look the same thing as before...same thing in both parenthesis.and 2 coefficients outside...so again
(x+4)(3x−2)
x5−3x3−2x2+6
We group the terms together by 2's
(x5−3x3)−(2x2−6)
Notice how that highlighted '-' is there...it WAS '+' but since we have a '-' in front of the parenthesis...when that distributes..it turns that '+' '-' again So now...we factor out of each parenthesis what we can... x^5 - 3x^3 ...well looks like at least x^3 can come out of there 2x^2 - 6 ...well we can factor a 2 out of there.. so now we have
x3(x2−3)−2(x2−3)
Now notice how we have 2 parenthesis that have the same thing inside them...and we also have 2 coefficients of those parenthesis...so we put those together respectively
(coefficients in here)(those things in the parentheis since they are the same)
(x3−2)(x2−3)
Everything make sense?
Well say we have
3x2+10x−8
We can actually break up that 10x ...and make it -2x + 12x right? since -2 + 12 =10 so we would then have
3x2−2x+12x−8
NOW we can group these up by 2's
(3x2−2x)+(12x−8)
Lets factor out an 'x' of the first set of parenthesis...and also lets factor out a 4 of the last set...
x(3x−2)+4(3x−2)
Look the same thing as before...same thing in both parenthesis.and 2 coefficients outside...so again
(x+4)(3x−2)
Lets suppose that the 4-term polynomial that is being factored is [tex]x^5-3x^2x^2+6[/tex].
[tex]x^5-3x^2x^2+6[/tex].
Factor out common terms in the first two terms, then in the last two terms
[tex]x^3(x^2-3)-2(x^2-3) [/tex]
Factor out the common term [tex]({x}^{2}-3)[/tex]
[tex]({x}^{2}-3)(x^3-2)[/tex]
Now lets factor a quadratic trinomial using this method
[tex]3x^2+10x-8[/tex]
Split the second term into two terms
Multiply the coefficient of the first term by the constant term
[tex]3*-8=-24 [/tex]
Ask yourself: Which two numbers will add up to 10 and multiply to -24?
[tex]12 \ \text{and} \ -2[/tex]
Split 10x as the sum of 12x and -2x
[tex]3x^2+12x-2x-8[/tex]
Factor out common terms in the first two terms, then in the last two terms
[tex]3x(x+4)-2(x+4)[/tex]
Factor out the common term x+4
[tex](x+4)(3x-2)[/tex]
[tex]x^5-3x^2x^2+6[/tex].
Factor out common terms in the first two terms, then in the last two terms
[tex]x^3(x^2-3)-2(x^2-3) [/tex]
Factor out the common term [tex]({x}^{2}-3)[/tex]
[tex]({x}^{2}-3)(x^3-2)[/tex]
Now lets factor a quadratic trinomial using this method
[tex]3x^2+10x-8[/tex]
Split the second term into two terms
Multiply the coefficient of the first term by the constant term
[tex]3*-8=-24 [/tex]
Ask yourself: Which two numbers will add up to 10 and multiply to -24?
[tex]12 \ \text{and} \ -2[/tex]
Split 10x as the sum of 12x and -2x
[tex]3x^2+12x-2x-8[/tex]
Factor out common terms in the first two terms, then in the last two terms
[tex]3x(x+4)-2(x+4)[/tex]
Factor out the common term x+4
[tex](x+4)(3x-2)[/tex]
