For any number c, when a polynomial p(x) of degree n≥ 1 is divided by (x-c), the remainder is p(c).
To divide the given polynomial we can use polynomial long division. When doing that all the terms of the dividend must be present and the polynomial must be in standard form.
p(x) = x^3-3x^2-7x+9
Since there are no missing terms and our polynomial is in descending degree order, we do not need to rewrite the polynomial. In our case, c=5. Therefore, let's divide the given polynomial by (x-5).
l r x - 5 & |l x^3-3x^2-7x+9
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Divide
x^3/x= x^2
r x^2 r x-5 & |l x^3-3x^2-7x+9
Multiply term by divisor
r x^2 rl x-5 & |l x^3-3x^2-7x+9 & x^3-5x^2
Subtract down
r x^2 r x-5 & |l 2x^2-7x+9
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Divide
2x^2/x= 2x
r x^2+2x r x-5 & |l 2x^2-7x+9
Multiply term by divisor
rx^2+ 2x rl x-5 & |l 2x^2-7x+9 & 2x^2-10x
Subtract down
r x^2+2x r x-5 & |l 3x+9
â–Ľ
Divide
3x/x= 3
r x^2+2x+3 r x-5 & |l 3x+9
Multiply term by divisor
rx^2+2x+ 3 rl x-5 & |l 3x+9 & 3x-15
Subtract down
r x^2+2x+3 r x-5 & |l 24
The quotient is x^2+2x+3 with a remainder of 24. Therefore, by the Remainder Theorem p(5)=24. Because we obtained the same result as in Part A, we also verified the theorem.
