Express 5x³+9x²–8x–11 in the form (x–1)(x+2)Q(x) + rx + s, find the values of r and s.
well, if x=1, the form (x-1)(X+2)Q(x) + rx + s, or equal to -r+s
so, -5+9+8-11=-r+s
same logic if x=-2, the new form becomes (-3)(O)(Q(-2) +r(-2)+s or
5((-8))+36-16-11=-2r+s
so this leads to two equations, two unknowns, r,s. Solve.
I did a long division:
(5x³+9x²–8x–11) ÷ (x^2 + 2 - 2) to get
5x³+9x²–8x–11 = (x-1)(x+2)(5x+3) - x - 5
comparing this to
(x-1)(x+2)Q(x) + rx + s
so rx = -x ---> r = -1
and s = -5
I looked at Bob's solution and noticed that in the 2nd and 4th lines
-5+9+8-11=-r+s should be 5+9-8-11= r+s -----> r + s = -5
-40+36-16-11 = -2r+s should be -40+36+16-11 = -2r+s ---> -2r+s = 1
solving for r and s gave me
r = -2 and s = -3
which is not the same as mine, how do you reconcile this??
I could only conclude that my Q(x) would have to be different than Bob's Q(x)
Bob would have:
5x³+9x²-8x-11 = (x-1)(x+2)Q(x) - 2x - 3 , or
5x³+9x²-6x-8 = (x-1)(x+2)Q(x)
I then divided 5x³+9x²-6x-8 by x^2 + x -2 and got Q(x) = 5x + 4
Ahhh, so both answers are correct
Bob's: 5x³+9x²-8x-11 = (x-1)(x+2)(5x+4) - 2x - 3
Mine: 5x³+9x²-8x-11 = (x-1)(x+2)(5x+3) - x - 5
and the Q(x)'s were indeed different
and there is no unique solution!
e.g.
60 = 2*3*8 + 12
60 = 2*3*9 + 6
To express the polynomial 5x³ + 9x² – 8x – 11 in the given form (x – 1)(x + 2)Q(x) + rx + s, we need to perform polynomial long division.
Step 1: Divide the first term of the polynomial by the first term of the divisor.
5x³ ÷ x = 5x²
Step 2: Multiply the divisor by the quotient obtained in Step 1.
(x – 1)(x + 2) = x² + 2x – x – 2 = x² + x – 2
Step 3: Subtract the result from Step 2 from the original polynomial.
(5x³ + 9x² – 8x – 11) – (x² + x – 2) = 5x³ + 9x² – 8x – 11 – x² – x + 2 = 5x³ + 8x² – 9x – 9
Step 4: Repeat Steps 1-3 until there are no more terms to divide.
5x³ ÷ x = 5x²
5x² ÷ x = 5x
(x – 1)(x + 2)Q(x) = (x² + x – 2)(5x² + 5x) = 5x⁴ + 10x³ + 5x³ + 5x² – 10x² – 2x² + 5x – 10x – 2x – 4 = 5x⁴ + 15x³ – 7x² – 7x – 4
Step 5: Subtract the result from Step 4 from the polynomial.
(5x³ + 8x² – 9x – 9) – (5x⁴ + 15x³ – 7x² – 7x – 4) = -5x⁴ – 7x³ + 15x² – 2x – 5
At this point, we have -5x⁴ – 7x³ + 15x² – 2x – 5 remaining.
Since we cannot divide x – 1 into this polynomial further, we can write the polynomial as:
(5x³ + 9x² – 8x – 11) = (x – 1)(x + 2)(5x² + 5x) + (-5x⁴ – 7x³ + 15x² – 2x – 5)
Therefore, r = -5 and s = -5.
To express the given polynomial 5x³+9x²–8x–11 in the desired form of (x–1)(x+2)Q(x) + rx + s, we need to perform polynomial division.
Step 1: Divide the given polynomial by (x–1)(x+2).
To do this, we set up the polynomial division as follows:
___________________________
(x–1)(x+2) | 5x³ + 9x² – 8x – 11
Now, start dividing by the highest power of x in the divisor, which is x².
Step 2: Divide 5x³ by x²:
___________
(x–1)(x+2) | 5x³ + 9x² – 8x – 11
- 5x³ + 5x²
___________
4x² – 8x – 11
Step 3: Bring down the next term, which is -8x:
___________
(x–1)(x+2) | 5x³ + 9x² – 8x – 11
- 5x³ + 5x²
___________
4x² – 8x – 11
- 4x² + 8x
_____________
0x – 11
Step 4: We have reached a remainder of 0x – 11, indicating that x–1 and x+2 are factors of the given polynomial.
Now, we can rewrite the given polynomial as:
5x³ + 9x² – 8x – 11 = (x–1)(x+2)Q(x) + rx + s
Since there is no remainder, we know Q(x) is equal to zero.
Therefore,
5x³ + 9x² – 8x – 11 = (x–1)(x+2)(0) + rx + s
= rx + s
Comparing the coefficients of x in the original polynomial and the rewritten form, we can find the values of r and s.
Coefficients of x in the original polynomial: -8
Coefficients of x in the rewritten form: r
Hence, r = -8.
Constant term in the original polynomial: -11
Constant term in the rewritten form: s
Hence, s = -11.
Therefore, the values of r and s are r = -8 and s = -11, respectively.