Respuesta :
Answer:
m = -6/7
Step-by-step explanation:
The constant term in the sum is ...
2·3 +7·m = 6+7m
In order for r(x) to be divisible by x, this value must be zero:
6 + 7m = 0
6/7 + m = 0 . . . . . divide by 7
m = -6/7 . . . . . . . . subtract 6/7
Answer:
[tex]m=\frac{-6}{7}[/tex]
Step-by-step explanation:
[tex]r(x)=2(x^2-11x+3)+7(x+m)[/tex]
[tex]m[/tex] is constant.
[tex]\frac{r(x)}{x}[/tex] has no remainder. This would mean [tex]x[/tex] is a factor of [tex]r[/tex].
By factor theorem if x-c is a factor of r then r(c)=0.
We have that x or x-0 is a factor of r:
By factor theorem if x-0 is a factor of r then r(0)=0.
So let's plug in 0 for x:
[tex]r(0)=2(0^2-11(0)+3)+7(0+m)[/tex]
[tex]r(0)=2(0-0+3)+7(m)[/tex]
[tex]r(0)=2(3)+7m[/tex]
[tex]r(0)=6+7m[/tex]
Now again we have r(0)=0 since x is a factor of r.
[tex]6+7m=0[/tex]
Subtract 6 on both sides:
[tex]7m=-6[/tex]
Divide both sides by 7:
[tex]m=\frac{-6}{7}[/tex]
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Another way.
So since [tex]r(x)[/tex] is divisible by [tex]x[/tex] that means when we divide [tex]r(x)[/tex] by [tex]x[/tex] we will have no fractions.
So let's do that:
[tex]\frac{2(x^2-11x+3)+7(x+m)}{x}[/tex]
[tex]\frac{2(x^2-11x+3)}{x}+\frac{7(x+m)}{x}[/tex]
[tex]2(\frac{x^2-11x+3}{x})+7(\frac{x+m}{x})[/tex]
[tex]2(\frac{x^2}{x}-\frac{11x}{x}+\frac{3}{x})+7(\frac{x}{x}+\frac{m}{x})[/tex]
[tex]2(x-11+\frac{3}{x})+7(1+\frac{m}{x})[/tex]
I'm going to get my fractions together.
Distribute the 2 to the terms in the ( ) next to it. Distribute 7 to the terms in the ( ) next to it:
[tex]2x-22+\frac{6}{x}+7+\frac{7m}{x}[/tex]
Reorder using commutative property of addition:
[tex]2x-22+7+\frac{6}{x}+\frac{7m}{x}[/tex]
Combine like terms ( the fractions too since the denominators are the same):
[tex]2x-15+\frac{6+7m}{x}[/tex]
Again we wanted no fraction.
A fraction is 0 when the numerator is 0.
If we find when 6+7m is 0, then we have found the value of m for which x divides r(x).
6+7m=0
Subtract 6 on both sides:
7m=-6
Divide both sides by 7:
m=-6/7.
