# Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the sameP(x) = 3x^4 - 5x^3 + x^2 - 3x + 5 Q(x) = (((3x - 5)x + 1)x - 3)x + 5

Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}-{5}{x}^{{3}}+{x}^{{2}}-{3}{x}+{5}\ {Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$ Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial $$\displaystyle{R}{\left({x}\right)}={x}^{{5}}—{2}{x}^{{4}}+{3}{x}^{{3}}—{2}{x}^{{3}}+{3}{x}+{4}$$ in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?

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Given

$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}−{5}{x}^{{3}}+{x}^{{2}}−{3}{x}+{5}$$

$${Q}{\left({x}\right)}={\left({\left({\left({3}{x}−{5}\right)}{x}+{1}\right)}{x}−{3}\right)}{x}+{5}$$

$${R}{\left({x}\right)}={x}^{{5}}−{2}{x}^{{4}}+{3}{x}^{{3}}−{2}{x}^{{2}}+{3}{x}+{4}$$

Expand Q

$$\displaystyle{Q}{\left({x}\right)}={\left({\left({\left({3}{x}−{5}\right)}{x}+{1}\right)}{x}−{3}\right)}{x}+{5}$$

$$={\left({\left({3}{x}^{{2}}−{5}{x}+{1}\right)}{x}−{3}\right)}{x}+{5}$$

$$={\left({3}{x}^{{3}}−{5}{x}^{{2}}+{x}−{3}\right)}{x}+{5}$$

$$={3}{x}^{{4}}−{5}{x}^{{3}}+{x}^{{2}}−{3}{x}+{5}$$

So, $$\displaystyle{P}{\left({x}\right)}={Q}{\left({x}\right)}={3}{x}^{{4}}−{5}{x}^{{3}}+{x}^{{2}}−{3}{x}+{5}$$

Hence proved Evaluate P(2) and Q(2)

$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}−{5}{x}^{{3}}+{x}^{{2}}−{3}{x}+{5}$$

$${P}{\left({2}\right)}={3}{\left({2}\right)}^{{4}}−{5}{\left({2}\right)}^{{3}}+{\left({2}\right)}^{{2}}−{3}{\left({2}\right)}+{5}$$

$$={48}−{40}+{4}−{6}+{5}$$

$$={11}$$

$${Q}{\left({2}\right)}={\left({\left({\left({3}{\left({2}\right)}−{5}\right)}{2}+{1}\right)}{2}−{3}\right)}{2}+{5}$$

$$={\left({\left({3}{\left({2}\right)}+{1}\right)}{2}−{3}\right)}{2}+{5}$$

$$={\left({\left({3}{\left({2}\right)}\right)}−{3}\right)}{2}+{5}$$

$$={\left({3}\right)}{2}+{5}$$

$$={11}$$
Nested form of R(x)

$$\displaystyle{R}{\left({x}\right)}={x}^{{5}}−{2}{x}^{{4}}+{3}{x}^{{3}}−{2}{x}^{{2}}+{3}{x}+{4}$$

$${R}{\left({x}\right)}={\left({x}^{{4}}−{2}{x}^{{3}}+{3}{x}^{{2}}−{2}{x}+{3}\right)}{x}+{4}$$

$$={\left({\left({x}^{{3}}−{2}{x}^{{2}}+{3}{x}−{2}\right)}{x}+{3}\right)}{x}+{4}$$

$$={\left({\left({\left({x}^{{2}}−{2}{x}+{3}\right)}{x}−{2}\right)}{x}+{3}\right)}{x}+{4}$$

$$={\left({\left({\left({\left({x}−{2}\right)}{x}+{3}\right)}{x}−{2}\right)}{x}+{3}\right)}{x}+{4}$$

$${R}{\left({x}\right)}={\left({\left({\left({\left({x}−{2}\right)}{x}+{3}\right)}{x}−{2}\right)}{x}+{3}\right)}{x}+{4}$$

$${R}{\left({3}\right)}={\left({\left({\left({\left({3}−{2}\right)}{3}+{3}\right)}{3}−{2}\right)}{3}+{3}\right)}{3}+{4}$$

$$={167}$$