Question

SAT Math (Thanks and please help!)

r(x) = 2(x^2-11x+3)+7(x+m)

In the polynomial defined above, m is a constant. If r(x) is divisible by x, what is m?

Answer 1

`m=(-6)/(7)`

Explanation:

`r(x)=2(x^2-11x+3)+7(x+m)`

`m` is constant.

`(r(x))/(x)` has no remainder. This would mean
`x` is a factor of
`r`.

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:

`r(0)=2(0^2-11(0)+3)+7(0+m)`

`r(0)=2(0-0+3)+7(m)`

`r(0)=2(3)+7m`

`r(0)=6+7m`

Now again we have r(0)=0 since x is a factor of r.

`6+7m=0`

Subtract 6 on both sides:

`7m=-6`

Divide both sides by 7:

`m=(-6)/(7)`

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Another way.

So since
`r(x)` is divisible by
`x` that means when we divide
`r(x)` by
`x` we will have no fractions.

So let's do that:

`(2(x^2-11x+3)+7(x+m))/(x)`

`(2(x^2-11x+3))/(x)+(7(x+m))/(x)`

`2((x^2-11x+3)/(x))+7((x+m)/(x))`

`2((x^2)/(x)-(11x)/(x)+(3)/(x))+7((x)/(x)+(m)/(x))`

`2(x-11+(3)/(x))+7(1+(m)/(x))`

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:

`2x-22+(6)/(x)+7+(7m)/(x)`

Reorder using commutative property of addition:

`2x-22+7+(6)/(x)+(7m)/(x)`

Combine like terms ( the fractions too since the denominators are the same):

`2x-15+(6+7m)/(x)`

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.

Answer 2

m = -6/7

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