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Sagot :
Answer:
The remainder is 3x - 4
Step-by-step explanation:
[Remember] [tex]\frac{Dividend}{Divisor} = Quotient + \frac{Remainder}{Divisor}[/tex]
So, [tex]Dividend = (Quotient)(Divisor) + Remainder[/tex]
In this case our dividend is always P(x).
Part 1
When the divisor is [tex](x - 1)[/tex], the remainder is [tex]4[/tex], so we can say [tex]P(x) = (Quotient)(x - 1) + 4[/tex]
In order to get rid of "Quotient" from our equation, we must multiply it by 0, so [tex](x - 1) = 0[/tex]
When solving for [tex]x[/tex], we get
[tex]x - 1 = 0\\x - 1 + 1 = 0 + 1\\x = 1[/tex]
When [tex]x = 1[/tex],
[tex]P(x) = (Quotient)(x - 1) + 4\\P(1) = (Quotient)(1 - 1) + 4\\P(1) = (Quotient)(0) + 4\\P(1) = 0 + 4\\P(1) = 4[/tex]
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Part 2
When the divisor is [tex](x + 3)[/tex], the remainder is [tex]104[/tex], so we can say [tex]P(x) = (Quotient)(x + 3) + 104[/tex]
In order to get rid of "Quotient" from our equation, we must multiply it by 0, so [tex](x + 3) = 0[/tex]
When solving for [tex]x[/tex], we get
[tex]x + 3 = 0\\x + 3 - 3 = 0 - 3\\x = -3[/tex]
When [tex]x = -3[/tex],
[tex]P(x) = (Quotient)(x + 3) + 104\\P(-3) = (Quotient)(-3 + 3) + 104\\P(-3) = (Quotient)(0) + 104\\P(-3) = 0 + 104\\P(-3) = 104[/tex]
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Part 3
When the divisor is [tex](x^2 - x + 1)[/tex], the quotient is [tex](x^2 + x + 3)[/tex], and the remainder is [tex](ax + b)[/tex], so we can say [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]
From Part 1, we know that [tex]P(1) = 4[/tex] , so we can substitute [tex]x = 1[/tex] and [tex]P(x) = 4[/tex] into [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]
When we do, we get:
[tex]4 = (1^2 + 1 + 3)(1^2 - 1 + 1) + a(1) + b\\4 = (1 + 1 + 3)(1 - 1 + 1) + a + b\\4 = (5)(1) + a + b\\4 = 5 + a + b\\4 - 5 = 5 - 5 + a + b\\-1 = a + b\\a + b = -1[/tex]
We will call [tex]a + b = -1[/tex] equation 1
From Part 2, we know that [tex]P(-3) = 104[/tex], so we can substitute [tex]x = -3[/tex] and [tex]P(x) = 104[/tex] into [tex]P(x) = (x^2 + x + 3)(x^2 - x + 1) + (ax + b)[/tex]
When we do, we get:
[tex]104 = ((-3)^2 + (-3) + 3)((-3)^2 - (-3) + 1) + a(-3) + b\\104 = (9 - 3 + 3)(9 + 3 + 1) - 3a + b\\104 = (9)(13) - 3a + b\\104 = 117 - 3a + b\\104 - 117 = 117 - 117 - 3a + b\\-13 = -3a + b\\(-13)(-1) = (-3a + b)(-1)\\13 = 3a - b\\3a - b = 13[/tex]
We will call [tex]3a - b = 13[/tex] equation 2
Now we can create a system of equations using equation 1 and equation 2
[tex]\left \{ {{a + b = -1} \atop {3a - b = 13}} \right.[/tex]
By adding both equations' right-hand sides together and both equations' left-hand sides together, we can eliminate [tex]b[/tex] and solve for [tex]a[/tex]
So equation 1 + equation 2:
[tex](a + b) + (3a - b) = -1 + 13\\a + b + 3a - b = -1 + 13\\a + 3a + b - b = -1 + 13\\4a = 12\\a = 3[/tex]
Now we can substitute [tex]a = 3[/tex] into either one of the equations, however, since equation 1 has less operations to deal with, we will use equation 1.
So substituting [tex]a = 3[/tex] into equation 1:
[tex]3 + b = -1\\3 - 3 + b = -1 - 3\\b = -4[/tex]
Now that we have both of the values for [tex]a[/tex] and [tex]b[/tex], we can substitute them into the expression for the remainder.
So substituting [tex]a = 3[/tex] and [tex]b = -4[/tex] into [tex]ax + b[/tex]:
[tex]ax + b\\= (3)x + (-4)\\= 3x - 4[/tex]
Therefore, the remainder is [tex]3x - 4[/tex].
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