Long Division Question

[Lil_yoyobo needs help]

Is (3a+1) a factor of P(x)= 3a^2 - 13a + 7 ?
Also an additional question is how to solve this using long division.
Help is appreciated.

 

[Steven G]

3a+1=0 when a=-1/3
(3a+1) is a factor of P(a) if and only if P(-1/3)=0. Is P(-1/3)=0?

Another way is to realize that only 7 works for 1*? =7. So if 3a+1 is a factor the other factor will need to be a+7. Is (3a+1)(a+7) = P(a)?

In the future to receive help on this forum you need to show your work so that the helpers here know what exactly you need help with

 

[JeffM]

Do you understand WHY [MATH]3a + 1[/MATH] is

a factor of [MATH]3a^2 - 13a + 7[/MATH] ONLY IF [MATH]3 * \left ( -\ \dfrac{1}{3} \right )^2 - 13 * \left (-\ \dfrac{1}{3} \right ) + 7 = 0.[/MATH]
As for long division, what is REALLY going on with long division?

We start with a simple division: [MATH]3a^2 \div 3a = a.[/MATH] So a is our first term.

Then we multiply [MATH]a \times (3a + 1) = 3a^2 + a.[/MATH]
Now we subtract [MATH](3a^2 - 13a) - (3a^2 + a) = -\ 14a.[/MATH]
And finally we "bring down" the 7 to get [MATH]-\ 14a + 7.[/MATH]
How does that make any sense? We are condensing the following logic.

[MATH]\dfrac{3a^2 - 13a + 7}{3a + 1} = \dfrac{3a^2 +0 -13a + 7}{3a + 1} = \dfrac{3a^2 +(a - a) - 13a + 7}{3a + 1} =[/MATH]
[MATH]\dfrac{(3a^2 + a) - a - 13a + 7}{3a + 1} = \dfrac{a(3a + 1) - 14a + 7}{3a + 1} = a + \dfrac{-\ 14a + 7}{3a + 1}.[/MATH]
So - 14/3 is our second term.

Now we repeat the process with a new numerator [MATH]-\ 14a + 7.[/MATH]
First, a simple division: [MATH]-\ 14a \div 3a = -\ \dfrac{14}{3}.[/MATH]
Next multiplication [MATH]-\ \dfrac{14}{3} * (3a + 1) = \left (-\ 14a - \dfrac{14}{3} \right ).[/MATH]
And then subtraction [MATH](-\ 14a + 7) - \left (-\ 14a - \dfrac{14}{3} \right) = 7 + \dfrac{14}{3} = \dfrac{35}{3}.[/MATH]
There is nothing left to "bring down," so we have a remainder of 35/3. (Going back to the original problem, a non-zero remainder means that (3a + 1) is not a factor.)

Again, what is the mathematical justification for the procedure?

[MATH]\dfrac{3a^2 - 13a + 7}{3a + 1} = a + \dfrac{-\ 14a + 7}{3a + 1} =[/MATH]
[MATH]a + \dfrac{-\ 14a + 0 + 7}{3a + 1} = a + \dfrac{-\ 14a + \left ( -\ \dfrac{14}{3} + \dfrac{14}{3} \right ) + 7}{3a + 1} =[/MATH]
[MATH]a + \dfrac{\left (-\ 14a - \dfrac{14}{3} \right ) + \left (\dfrac{14}{3} + 7 \right )}{3a + 1} = a + \dfrac{\left (-\ \dfrac{14 * 3a}{3} - \dfrac{14 * 1}{3} \right ) + \dfrac{35}{3}}{3a + 1} =[/MATH]
[MATH]a + \dfrac{-\ \dfrac{14}{3} * (3a + 1) + \dfrac{35}{3}}{3a + 1} = a - \dfrac{14}{3} + \dfrac{35}{3(3a + 1)}.[/MATH]
Now we can check all this if you like.

[MATH](3a + 1) * \left (a - \dfrac{14}{3} + \dfrac{35}{3(3a + 1)}. \right ) = 3a^2 - 14a + a - \dfrac{14}{3} + \dfrac{35}{3} = [/MATH]
[MATH]3a^2 - 13a + \dfrac{35 - 14}{3} = 3a^2 - 13a + 7.[/MATH]

 

[HallsofIvy]

Is (3a+1) a factor of P(x)= 3a^2 - 13a + 7 ?
Also an additional question is how to solve this using long division.
Help is appreciated.

3a divides into \(\displaystyle 3a^2\) a times. 1 divides into 7 seven times. If 3a+ 1 is a factor of 3a^2- 13a+ 7 then the other factor would have to be a+ 7. (3a+ 1)(a+ 7)= 3a(a+ 7)+ 1(a+ 7)= 3a^2+ 21a+ a+ 7= 3a^2+ 22a+ 7, not 3a^2- 13a+ 7. No, 3a+ 1 is not a factor of 3a^2- 13a+ 7.

3a divides into \(\displaystyle 3a^2\) a times. Multiply a times 3a+1 to get \(\displaystyle 3a^2+ a\). Subtract from \(\displaystyle 3a^2- 13a+ 7\) to get -14a+ 7. 3a divides into -14a -14/3 times. -14/3 times 3a+ 1 is -14a+ 14/3. Subtract from -14a+ 7 to get 7/3.

3a+ 1 divides into 3a^2-