Divisibility Question!

[bobrossu]

Here's a problem:

If 2x³+x²+ax+b is divisible by x²+x-7, find a-b

I divided 2x³+x²+ax+b by x²+x-7 and gotten a remainder of (ax-13x)+(b+7).
Not sure what to do next.

Please show work so I can learn!

 

[tkhunny]

It is your work that must be shown. Better try that division again. You're close.

If it IS divisible, what IS the remainder?

Example: 12 / 4 = 3 Remainder 0
Example: 12/5 = 2 Remainder 2 <== Not 0

 

[bobrossu]

It is your work that must be shown. Better try that division again. You're close.

If it IS divisible, what IS the remainder?

Example: 12 / 4 = 3 Remainder 0
Example: 12/5 = 2 Remainder 2 <== Not 0

Checked my work and got (ax+15x)+(b-7) , but i'm still not sure what to do next.

 

[Deleted member 4993]

Checked my work and got (ax+15x)+(b-7) , but i'm still not sure what to do next.

I think you are forgetting long division!

If 2x³+x²+ax+b is divisible by x²+x-7, find a-b

You should get something like:

2x³+x²+ax+b = (p*x + q) * (x²+x-7) + (u * x + v)

What do you get for p, q, u & v?

 

[bobrossu]

I think you are forgetting long division!

If 2x³+x²+ax+b is divisible by x²+x-7, find a-b

You should get something like:

2x³+x²+ax+b = (p*x + q) * (x²+x-7) + (u * x + v)

What do you get for p, q, u & v?

(2x-1)*(x²-x-7)+(ax+15x+b-7)

trying to find u and v

 

[mmm4444bot]

(2x-1)*(x² - x -7 )+(ax+15x+b-7)

trying to find u and v

Factor ax + 15x. You'll then see the quantity multiplying x as a binomial expression.

As b is a constant, think of the binomial expression b - 7 as a single quantity (constant). :cool:

 

[tkhunny]

Your division is close enough - although more care is warranted. We need to discuss only the numerator of the Remainder portion.

You are missing but one thing...

In order for a polynomial to BE Zero (0), ALL of its coefficients must be Zero (0). This is what you need.

Example: If ax^2 + bx + c is the Constant Zero (0), for all values of x, then necessarily a = 0 and b = 0 and c = 0.

 

[bobrossu]

Your division is close enough - although more care is warranted. We need to discuss only the numerator of the Remainder portion.

You are missing but one thing...

In order for a polynomial to BE Zero (0), ALL of its coefficients must be Zero (0). This is what you need.

Example: If ax^2 + bx + c is the Constant Zero (0), for all values of x, then necessarily a = 0 and b = 0 and c = 0.

Ok ok, ! think I get it. So since it is divisible, then there should be no remainder.

So I just set the individual remainders equal to zero and solve for x.

x(a+15)=0
b-7=0

a=-15
b=7

 

[tkhunny]

Ok ok, ! think I get it. So since it is divisible, then there should be no remainder.

So I just set the individual remainders equal to zero and solve for x.

Not "individual remainders", but "individual coefficients". I think you have it! Good work.

You will need this same idea when trying to make two polynomials identical. Each corresponding coefficient must be equal.