After half a page of a messy long division of ax^3+5x^2+bx-2 by 2x^2+3x-2
I had a remainder of (b+a - (30-9a)/4)x -2+(10a-3a)/2
matching this with 9x - 6
we get b+a - (30-9a)/4 = 9 and -2+(10a-3a)/2 = -6
I will leave it up to you to solve for a and b
Hint: it comes out to "nice" integers
When the Polynomial p(x)=ax^3+5x^2+bx-2 where a and b are constants is divided by 2x^2+3x-2, the remainder is 9x-6.
(I) Find the values of a and b
(II) Using these values of a and b, factorize p(x) completely
(III) Find the remainder when p(x) is divided by (3x+1).
4 answers
Let p(x) = (2x^2+3x-2)(mx+c)+(9x-6)
If you expand that out, you get
2mx^3 + (3m+2c)x^2 + (3c-2m+9)x - (2c+6) = ax^2 + 5x^2 + bx - 2
That means you just have to solve the equations
2m = a
3m+2c = 5
3c-2m+9 = b
2c+6 = 2
Looks daunting, but if you start at the bottom, you get a value for c right away, and the rest falls out easily.
If you expand that out, you get
2mx^3 + (3m+2c)x^2 + (3c-2m+9)x - (2c+6) = ax^2 + 5x^2 + bx - 2
That means you just have to solve the equations
2m = a
3m+2c = 5
3c-2m+9 = b
2c+6 = 2
Looks daunting, but if you start at the bottom, you get a value for c right away, and the rest falls out easily.
Please its an assignment
It is not an assignment
1 answer