15 Polynomials
An expression of the form \(a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0\) where \(a_k,\) \(k=0,1,2,\cdots n\) are real numbers and \(a_n\neq 0\) is called a polynomial of degree
\(n\).
Polynomials like integers can be divided to get a quotient and a remainder.
- Denote a polynomial by \(P(x)\); \(p(x)\)
- Denote a quotient by \(Q(x)\); \(q(x)\)
- Denote a remainder by \(R(x)\); \(r(x)\)
Find the quotient and the remainder when the polynomial \(P(x)=x^3-7x^2+6x-2\) is divided by \(x-2\).
\[\begin {array}{r@{\;}l} & x^{2}-5x-4\\ x-2\,\big ) & \overline {x^{3}-7x^{2}+6x-2}\\ & \underline {x^{3}-2x^{2}}\phantom {{}+6x-2}\\ & \phantom {x^{3}}-5x^{2}+6x\phantom {{}-2}\\ & \phantom {x^{3}}\underline {-5x^{2}+10x}\phantom {{}-2}\\ & \phantom {x^{3}-5x^{2}}-4x-2\\ & \phantom {x^{3}-5x^{2}}\underline {-4x+8}\\ & \phantom {x^{3}-5x^{2}}-10 \end {array}\] The quotient is \(x^{2}-5x-4\) and the remainder is \(-10\).
Therefore \(Q(x)=x^2-5x-4\) and \(R(x)=-10\)
Solution.
Note that \(x^4-3\) can be written as \(x^4+0x^3+0x^2+0x-3\). Then we have \[\begin {array}{r@{\;}l} & x^{3}-2x^{2}+4x-8\\ x+2\,\big ) & \overline {x^{4}+0x^{3}+0x^{2}+0x-3}\\ & \underline {x^{4}+2x^{3}}\phantom {{}+0x^{2}+0x-3}\\ & \phantom {x^{4}}-2x^{3}+0x^{2}\phantom {{}+0x-3}\\ & \phantom {x^{4}}\underline {-2x^{3}-4x^{2}}\phantom {{}+0x-3}\\ & \phantom {x^{4}-2x^{3}}4x^{2}+0x\phantom {{}-3}\\ & \phantom {x^{4}-2x^{3}}\underline {4x^{2}+8x}\phantom {{}-3}\\ & \phantom {x^{4}-2x^{3}+4x^{2}}-8x-3\\ & \phantom {x^{4}-2x^{3}+4x^{2}}\underline {-8x-16}\\ & \phantom {x^{4}-2x^{3}+4x^{2}}13 \end {array}\] The quotient is \(x^{3}-2x^{2}+4x-8\) and the remainder is \(13\). The missing powers are written as \(0x^{3}\), \(0x^{2}\) and \(0x\) so that like terms stay in their own columns — leaving them out is the commonest cause of a slipped line. Therefore
- Quotient: \(Q(x)=x^3-2x^2+4x-8\)
- Remainder: \(R(x)=13\)
Solution. Writing the dividend in full as \(x^{3}+0x^{2}+0x-1\) and dividing gives \[Q(x)=x^{2}+x+1,\qquad R=0 .\] The remainder is zero, so \(x-1\) is a factor and \[x^{3}-1=(x-1)\left (x^{2}+x+1\right ),\] which is the standard difference-of-cubes factorisation. The remainder theorem below gives the same answer instantly: \(P(1)=1-1=0\).
In general if \(P(x)\) is a polynomial function then we divide \(P(x)\) by a linear factor \(x-a\) for some real number \(a\) we can
express \(P(x)\) in the form
\[P(x)=(x-a)Q(x)+R(x)\]
where \(Q(x)\) is the quotient and \(R(x)\) is the remainder.
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