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# help asap, thanks

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Let \(P(x)\) be a cubic polynomial such that \(P(0) = -3\) and \(P(1) = 4\) . When \(P(x)\) is divided by \(x^2 + x + 1\) the remainder is \(2x-1\) . What is the quotient when \(P(x)\) is divided by \(x^2+x+1\)?

Feb 18, 2020

#1
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Being a cubic polynomial, it has the form:  P(x)  =  ax3 + bx2 + cx + d

Since P(0) = -3   --->   P(0)  =  a(0)3 + b(0)2 + c(0) + d  =  -3   --->   d = -3

Since P(1) = 4   --->  P(1)  =  a(1)3 + b(1)2 + c(1) + -3  =  a + b + c - 3  =  4   --->   a + b + c  =  7

Using long division, I divided  ax3 + bx2 + cx - 3  by  x2 + x + 1  and I got

a quotient of  ax + (b - a)  and  a remainder of  (c - b)x + (a - b - 3)

Since the problem said that the remainder was  2x - 1   --->   (c - b)x + (a - b - 3)  =  2x - 1

which means that  c - b  =  2  and  a - b - 3  =  -1    --->   c  =  b + 2   and   a  =  b + 2

Combining  a + b + c  =  7  and  c = b + 2  and  a = b + 2

--->   a = 3   and   b = 1   and  c = 3

Since the quotiet was  ax + (b - a)  and  a = 3  and  b = 1,

Feb 19, 2020

#1
+1

Being a cubic polynomial, it has the form:  P(x)  =  ax3 + bx2 + cx + d

Since P(0) = -3   --->   P(0)  =  a(0)3 + b(0)2 + c(0) + d  =  -3   --->   d = -3

Since P(1) = 4   --->  P(1)  =  a(1)3 + b(1)2 + c(1) + -3  =  a + b + c - 3  =  4   --->   a + b + c  =  7

Using long division, I divided  ax3 + bx2 + cx - 3  by  x2 + x + 1  and I got

a quotient of  ax + (b - a)  and  a remainder of  (c - b)x + (a - b - 3)

Since the problem said that the remainder was  2x - 1   --->   (c - b)x + (a - b - 3)  =  2x - 1

which means that  c - b  =  2  and  a - b - 3  =  -1    --->   c  =  b + 2   and   a  =  b + 2

Combining  a + b + c  =  7  and  c = b + 2  and  a = b + 2

--->   a = 3   and   b = 1   and  c = 3

Since the quotiet was  ax + (b - a)  and  a = 3  and  b = 1,   