Show that the product of two monic polynomials is also a monic polynomial.
Answers
Answer:
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Step-by-step explanation:
Monic Polynomial
Definition. The polynomials in only one variable with leading coefficient 11 , are called monic polynomials.
Solution:
Let two monic polynomials be
\quad f(x)=x^{m}+c_{1}x^{m-1}+c_{2}x^{m-2}+...+c_{m-1}f(x)=x
m
+c
1
x
m−1
+c
2
x
m−2
+...+c
m−1
\quad g(x)=x^{n}+d_{1}x^{n-1}+d_{2}x^{m-2}+...+d_{n-1}g(x)=x
n
+d
1
x
n−1
+d
2
x
m−2
+...+d
n−1
where c_{i},\:d_{j}\in\mathbb{R}\:\text{or}\:\in\mathbb{C}c
i
,d
j
∈Ror∈C
If the products of the polynomials f(x)f(x) and g(x)g(x) be defined by h(x)h(x) , we can write
\quad h(x)=f(x)\times g(x)h(x)=f(x)×g(x)
\Rightarrow h(x)=(x^{m}+c_{1}x^{m-1}+...+c_{m-1})(x^{n}+c_{1}x^{n-1}+...+d_{n-1}⇒h(x)=(x
m
+c
1
x
m−1
+...+c
m−1
)(x
n
+c
1
x
n−1
+...+d
n−1
\Rightarrow h(x)=x^{m+n}+c_{1}x^{m+n-1}...+c_{m-1}d_{n-1}⇒h(x)=x
m+n
+c
1
x
m+n−1
...+c
m−1
d
n−1
Here h(x)h(x) has only one variable and the leading coefficient is 11 , and thus h(x)h(x) is also monic.
This completes the proof.
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3
+3x
2
−3 and Q(x) = 2x^{3} - 5x +kQ(x)=2x
3
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