HELP HELP HELP !

[Madisonmackk]

Okay so, i know i just went along and posted a math 11 thing about 3 seconds ago..
but I FOUND SOMETHING ELSE THATS CRAZY HARD !

Question is for example :

Which polynomial has x-5 as a factor ?

a) x[sup:11jowe22]3[/sup:11jowe22]-5x[sup:11jowe22]2[/sup:11jowe22]+2x-5

b) -x[sup:11jowe22]3[/sup:11jowe22]+5x[sup:11jowe22]2[/sup:11jowe22]-2x+7

c) x[sup:11jowe22]3[/sup:11jowe22]-10x[sup:11jowe22]2[/sup:11jowe22]+31x-30

I jsut can't figure out how to start this question without having to plug in (+5) in every single equation.. the answer is c, But why ?

 

[fasteddie65]

Unless you can factor mentally, which is quite challenging, substituting x = 5 is the best method. However you can use Synthetic Substitution rather than direct substitution.

 

[Deleted member 4993]

Okay so, i know i just went along and posted a math 11 thing about 3 seconds ago..
but I FOUND SOMETHING ELSE THATS CRAZY HARD !

Question is for example :

Which polynomial has x-5 as a factor ?

a) x[sup:2b3rrsyo]3[/sup:2b3rrsyo]-5x[sup:2b3rrsyo]2[/sup:2b3rrsyo]+2x-5

b) -x[sup:2b3rrsyo]3[/sup:2b3rrsyo]+5x[sup:2b3rrsyo]2[/sup:2b3rrsyo]-2x+7

c) x[sup:2b3rrsyo]3[/sup:2b3rrsyo]-10x[sup:2b3rrsyo]2[/sup:2b3rrsyo]+31x-30

I jsut can't figure out how to start this question without having to plug in (+5) in every single equation.. the answer is c, But why ?

That is the FASTEST way to solve this problem.

 

[soroban]

Hello, Madisonmackk!

Which polynomial has \(\displaystyle x-5\) as a factor ?

. . \(\displaystyle \begin{array}{cc}a)& x^3 - 5x^2+2x-5 \\ b)& -x^3 + 5x^2 -2x+7 \\ c) & x^3 -10x^2 +31x-30 \end{array}\)

I just can't figure out how to start this question without having to plug in (+5) in every single equation.
What wrong with that?
The alternative is to divide the three polynomials by (x-5).

The answer is c, But why ?
Um ... because it works?

Note that we can immediately eliminate (b) . . . It ends in 7.
Do you get the reasoning?