Solve Polynomial Division: Find K Given Remainder of 3

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To find the value of k in the polynomial 4x^2 + 9x + k when divided by x-1 with a remainder of 3, evaluate the polynomial at x = 1. By substituting x = 1 into the polynomial, you get 4(1)^2 + 9(1) + k, which simplifies to 13 + k. Set this equal to the given remainder of 3, resulting in the equation 13 + k = 3. Solving for k yields k = -10. This method eliminates the need for polynomial long division.
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Homework Statement


The remainder is 3 when 4x^2 + 9x + k is divided by x-1
Im suppose to find the K. I am just confused because this question they give me the remainder..so what am i suppose to do with it?
 
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Perform the usual operation, continue bringing k down and then at the end just solve for k by setting it equal to what your remainder should be.
 
thank you
 
You could also use the fact that when a polynomial, P(x), is divided by x- a, the remainder is exactly P(a). You don't need to do any division at all.
 

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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