Divide Polynomial Using Long Division: (6x^3-16x^2+11x-5) / (3x-2)?
….….… 2x² - 4x + 1 ------------------------------------- 3x - 2 ) 6x³ - 16x² + 11x - 5 ….… - 6x³ + 4x² ….… --------------- ….…….… - 12x² + 11x ….…..… + 12x² - 8x ….….…. --------------- ….……..…… + 3x - 5 ….……...…… - 3x + 2 ……........…. --------------- ….……..……..… - 3 ANSWER: (6x³ - 16x² + 11x - 5) / (3x - 2) = 2x² - 4x + 1 R -3 CHECK: (3x - 2)(2x² - 4x + 1) - 3 = 6x³ - 16x² + 11x - 5? 3x(2x²) + 3x(-4x) + 3x(1) - 2(2x²) - 2(-4x) - 2(1) - 3 = 6x³ - 16x² + 11x - 5? 6x³ - 12x² + 3x - 4x² + 8x - 2 - 3 = 6x³ - 16x² + 11x - 5? 6x³ - 16x² + 11 - 5 = 6x³ - 16x² + 11x - 5? true
Use polynomial long division to divide?
The first step is to set up your long division: +-------------------------- -3x^2 - 5x - 2 | -9x^3 -18x^2 -11x -2 To determine the first value to use, divide the first term of the numerator (the numerator is top number) by the highest term of the denominator (the bottom number). In this problem that is -9x^3/-3x^2 = 3x. Put 3x above the first term of the numerator and then multiply the denominator by 3x and write it underneath the numerator, like this: 3x +-------------------------------- -3x^2 - 5x - 2 | -9x^3 -18x^2 -11x -2 -9x^3 -15x^2 -6x Now subtract this from the numerator --------------------------- -3x^2 -5x Put the remainder of the subtraction here. Now we take the first term of the remainder (-3x^2) and divide it by the highest term of the denominator (-3x^2), and get -3x^2/-3x^2 = 1. We put the result (1) above the -18x^2, multiply the denominator by 1 write the result underneath the remainder, like this: 3x + 1 +-------------------------------- -3x^2 - 5x - 2 | -9x^3 -18x^2 -11x -2 -9x^3 -15x^2 -6x --------------------------- -3x^2 -5x -2 We bring the -2 down from the numerator -3x^2 -5x -2 Now subtract this from the remainder -------------------- 0 This is our new remainder. This tells us that: (-9x^3 -18x^2 -11x -2)/(-3x^2 - 5x - 2) = 3x + 1
Help with long division of polynomials?
Draw the |````````````` thing that you use in long division of numbers. Write the polynomial in the numerator inside the thing as 2x^5 - 7x^4 + 0x^3 + 0x^2 + 0x - 13. Then on the outside left write the polynomial in the denominator as 4x^2 - 6x + 8. The first term of the new polynomial is the ratio of the polynomial of highest degree inside the thing / outside the thing. Here it is 2x^5 / 4x^2 = (1/2)x^3. Just like long division of numbers. Multiply (1/2)x^3 by 4x^2 - 6x + 8 and subtract it from 2x^5 - 7x^4 + 0x^3 + 0x^2 + 0x - 13. So we get (1/2)x^3 * [4x^2 - 6x + 8] = 2x^5 - 3x^4 + 4x^3. So [2x^5 - 7x^4 + 0x^3 + 0x^2 + 0x - 13] - [2x^5 - 3x^4 + 4x^3] = -4x^4 - 4x^3 + 0x^2 + 0x - 13. Now we start the process over again. Now we compare the -4x^4 to the 4x^2 so -4x^4 / 4x^2 = -x^2 (this is the next term of our answer) and (-x^2)*[4x^2 - 6x + 8] = -4x^4 + 6x^3 - 8x^2. Now subtract this from our remaining polynomial to get: [-4x^4 - 4x^3 + 0x^2 + 0x - 13] - [-4x^4 + 6x^3 - 8x^2] = -10x^3 + 8x^2 + 0x - 13. Then -10x^3 / 4x^2 = (-5/2)x (this is the next term of our answer) and (-5/2)x * (4x^2 - 6x + 8) = - 10x^3 + 15x^2 - 20x. Now subtract this from our remaining polynomial to get: [-10x^3 + 8x^2 + 0x - 13] - [-10x^3 + 15x^2 - 20x] = -7x^2 + 20x - 13. So -7x^2 / 4x^2 = (-7/4) (this is the next term of our answer) and (-7/4)*(4x^2 - 6x + 8) = -7x^2 + (21/2)x - 14. Now subtract this from our remaining polynomial to get: [-7x^2 + 20x - 13] - [-7x^2 + (21/2)x - 14] = 19x/2 + 1. Now the degree of (19 x / 2 + 1) < degree (4x^2 - 6x + 8) so we can stop. By collection our terms together we get the new polynomial to be: (1/2)x^3 - x^2 - (5/2)x - (7/4) But we have a remainder and we write that as Whats left over / Demoniator of original polynoimal ((19/2)x + 1) / (x^2 - 6x + 8) So our answer is (1/2)x^3 - x^2 - (5/2)x - (7/4) + ((19/2)x + 1) / (x^2 - 6x + 8)
The polynomial p(x) =x^4-2x^3+3x^2-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19, respectively. What are a and b?
We have,p(x)=x^4–2x^3+3x^2-ax+bBy remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.By the given condition, we havep(1)=5 and p(-1)=19=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19=> -a+b=5–1+2–3 and 1+2+3+a+b=19=> -a+b=3 and a+b=19–1–2–3=> -a+b=3 and a+b=13Adding these two equations,we get-a+b+a+b=3+13=> 2b=16=> 2b/2=16/2=> b=8Putting b=8 in a+b=13 , we geta+8=13=> a=13–8=> a=5Therefore, a=5 and b=8 .
How do you divide polynomials by synthetic division if the divisior is (3x-2)?
3x^3 = x^2(3x-2)+2x^2 2x^2+4x^2 = 6x^2 6x^2 = 2x(3x-2)+4x 4x-7x = -3x -3x = -(3x-2) -2 -2+1=-1 3x^3+4x^2-7x+1 =x^2(3x-2) +2x(3x-2) -1(3x-2) -1 (3x^3+4x^2-7x+1)/(3x-2) = x^2 +2x -1 -1/(3x-2)
Polynomial Long Division help please?
Really need help with what I've done wrong. p(x)=(x+1)(x^2-4x+5) Find the remainder when p(x) is divided by x-2. What I did... (x+1)(x^2-4x+5) = x^3-3x^2+x+5 (which is apparently right) ......_____x^2+x+3___ x-2 | x^3-3x^2+x+5 .....-(x^3-2x^2)..|...| ..............x^2+x....| ............-(x^2-2x)..| ....................3x+5 ...................-(3x-6) ........................11 .....so remainder is 11 right? What the mark scheme did... p(2)=(2+1)(2^2-8+5) remainder = 3 _______________ wtf??? please explain!! Thank you.
Divide the following polynomial 6x^3+11x^2-4x-4/3x-2?
You have to use a method called polynomial long division. I found this question divides with no remainder, My answer was 2x^2+5x+2. - This shows a fairly clear way how to do this calculation: http://en.wikipedia.org/wiki/Polynomial_long_division