Question

# Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q are the same P(x)=3x^{4}-5x^{3}+x^{2}-3x+5 Q(x)=(((3x-5)x+1)x-3)x+5. Try to evalue P(2) and Q(2) in your head, using the forms given.

Polynomial arithmetic
Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q are the same.
$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{4}}}-{5}{x}^{{{3}}}+{x}^{{{2}}}-{3}{x}+{5}$$
$$\displaystyle{Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$
Try to evalue P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polinomial $$\displaystyle{R}{\left({x}\right)}={x}^{{{5}}}-{2}{x}^{{{4}}}+{3}{x}^{{{3}}}-{2}{x}^{{{2}}}+{3}{x}+{4}$$ in "nested" form, like the polinomial Q. Use the nested form to find R(3) in your head.

2021-07-30
Q(2) can be easily calculated (mentally) than P(2) since in each step we have to perform linear operations starting from the inside.
While evaluating P(2) mentally we find that larger number occurs in every step.
To find P(2):
$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{4}}}-{5}{x}^{{{3}{\mid}+{x}^{{{2}}}-{3}{x}+{5}}}$$
$$\displaystyle{P}{\left({2}\right)}={3}{\left({2}\right)}^{{{4}}}-{5}{\left({2}\right)}^{{{3}}}+{\left({2}\right)}^{{{2}}}-{3}{\left({2}\right)}+{5}$$
$$\displaystyle={3}{\left({16}\right)}-{5}{\left({8}\right)}+{4}-{6}+{5}$$
$$\displaystyle={48}-{40}+{4}-{6}+{5}$$
$$\displaystyle={57}-{46}$$
$$\displaystyle={11}$$
To find Q(2):
$$\displaystyle{Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$
$$\displaystyle{Q}{\left({2}\right)}={\left({\left({\left({3}{\left({2}\right)}-{5}\right)}{2}+{1}\right)}{2}-{3}\right)}{2}+{5}$$
$$\displaystyle={\left({\left({\left({1}\right)}{2}+{1}\right)}{2}-{3}\right)}{2}+{5}$$
$$\displaystyle={\left({\left({3}\right)}{2}-{3}\right)}{2}+{5}$$
$$\displaystyle={\left({6}-{3}\right)}{2}+{5}$$
$$\displaystyle={3}{\left({2}\right)}+{5}$$
$$\displaystyle={6}+{5}$$
$$\displaystyle={11}$$
Now we have to write R(x) in nested form.
$$\displaystyle{R}{\left({x}\right)}={x}^{{{5}}}-{2}{x}^{{{4}}}+{3}{x}^{{{3}}}-{2}{x}^{{{2}}}+{3}{x}+{4}$$
Now we have factor out x.
$$\displaystyle{R}{\left({x}\right)}={\left({x}^{{{4}}}-{2}{x}^{{{3}}}+{3}{x}^{{{2}}}-{2}{x}+{3}\right)}{x}+{4}$$
$$\displaystyle={\left({\left({x}^{{{3}}}-{2}{x}^{{{2}}}+{3}{x}-{2}\right)}{x}+{3}\right)}{x}+{4}$$
$$\displaystyle={\left({\left({\left({x}^{{{2}}}-{2}{x}+{3}\right)}{x}-{2}\right)}{x}+{3}\right)}{x}+{4}$$
$$\displaystyle={\left({\left({\left({\left({x}-{2}\right)}{x}+{3}\right)}{x}-{2}\right)}{x}+{3}\right)}{x}+{4}$$
We can calculate R(3) mentally.
$$\displaystyle{R}{\left({x}\right)}={\left({\left({\left({\left({x}-{2}\right)}{x}+{3}\right)}{x}-{2}\right)}{x}+{3}\right)}{x}+{4}$$
$$\displaystyle{R}{\left({3}\right)}={\left({\left({\left({\left({3}-{2}\right)}{3}+{3}\right)}{3}-{2}\right)}{3}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({\left({\left({\left({1}\right)}{3}+{3}\right)}{3}-{2}\right)}{3}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({\left({\left({6}\right)}{3}-{2}\right)}{3}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({\left({18}-{2}\right)}{3}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({\left({16}\right)}{3}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({48}+{3}\right)}{3}+{4}$$
$$\displaystyle={\left({51}\right)}{3}+{4}$$
$$\displaystyle={153}+{4}$$
$$\displaystyle={157}$$
$$\displaystyle{R}{\left({3}\right)}={157}$$
The polynomials P and Q are the same by expanding Q is proved.