Q6. Find
2/3x-9/8x7/4
Answers
Answer:
Step-by-step explanation:
The polynomial {\displaystyle 3-5x+2x^{5}-7x^{9}} 3-5x+2x^{5}-7x^{9} is a nonic polynomial
The polynomial {\displaystyle (y-3)(2y+6)(-4y-21)} (y-3)(2y+6)(-4y-21) is a cubic polynomial
The polynomial {\displaystyle (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z)} (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z) is a quintic polynomial (as the {\displaystyle z^{8}} z^{8} are cancelled out)
The canonical forms of the three examples above are:
for {\displaystyle 3-5x+2x^{5}-7x^{9}} 3-5x+2x^{5}-7x^{9}, after reordering, {\displaystyle -7x^{9}+2x^{5}-5x+3} -7x^{9}+2x^{5}-5x+3;
for {\displaystyle (y-3)(2y+6)(-4y-21)} (y-3)(2y+6)(-4y-21), after multiplying out and collecting terms of the same degree, {\displaystyle -8y^{3}-42y^{2}+72y+378} -8y^{3}-42y^{2}+72y+378;
for {\displaystyle (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z)} (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z), in which the two terms of degree 8 cancel, {\displaystyle z^{5}+8z^{4}+2z^{3}-4z^{2}+14z+6} z^{5}+8z^{4}+2z^{3}-4z^{2}+14z+6
Step-by-step explanation:
The degree of {\displaystyle (x^{3}+x)+(x^{2}+1)=x^{3}+x^{2}+x+1} (x^{3}+x)+(x^{2}+1)=x^{3}+x^{2}+x+1 is 3. Note that 3 ≤ max(3, 2)
The degree of {\displaystyle (x^{3}+x)-(x^{3}+x^{2})=-x^{2}+x} (x^{3}+x)-(x^{3}+x^{2})=-x^{2}+x is 2. Note that 2 ≤ max(3, 3)
Scalar multiplication
Edit
The degree of the product of a polynomial by a non-zero scalar is equal to the degree of the polynomial, i.e.
{\displaystyle \deg(cP)=\deg(P)} \deg(cP)=\deg(P).
E.g. The degree of {\displaystyle 2(x^{2}+3x-2)=2x^{2}+6x-4} 2(x^{2}+3x-2)=2x^{2}+6x-4 is 2, just as the degree of {\displaystyle x^{2}+3x-2} x^{2}+3x-2.
Note that for polynomials over a ring containing divisors of zero, this is not necessarily true. For example, in {\displaystyle \mathbf {Z} /4\mathbf {Z} } {\mathbf {Z}}/4{\mathbf {Z}}, {\displaystyle \deg(1+2x)=1} \deg(1+2x)=1, but {\displaystyle \deg(2(1+2x))=\deg(2+4x)=\deg(2)=0} \deg(2(1+2x))=\deg(2+4x)=\deg(2)=0.