Dolly Robinson

2020-10-28

(a)To calculate: The following equation $\left[\left(3x+5\right)x+4\right]x+3x+1=3{x}^{4}+5{x}^{3}+4{x}^{2}+3x+1$ is an identity, (b) To calculate: The lopynomial $P\left(x\right)=6{x}^{5}-3{x}^{4}+9{x}^{3}+6{x}^{2}-8x+12$ without powers of x as in patr (a).

### Answer & Explanation

mhalmantus

Given: A fourth-degree polynomial in x such as $3{x}^{4}+5{x}^{3}+4{x}^{2}+3x+1$ contains all of the powers of x from the first through the fourth. However, any polynomial can be written without powers of x. Evaluating a polynomial without powers of x (Horner's method) is somewhat easier than evaluating a polynomial with powers. Formula used: Expansion of a polynomial, Suppose, an equation is given in the form $a\left(x+2\right)x$ can be expanded and written as, $a\left(x+2\right)x=a{x}^{2}+2ax.$ This is known as expansion of an equation. Calculation: The given equation $\left[\left(3x+5\right)x+4\right]x+3x+1$ can be expanded as follow, $\left[\left(3x+5\right)x+4\right]x+3x+1=\left[3{x}^{2}+5x+4\right]x+3x+1$
$=3{x}^{3}+5{x}^{2}+4x+3x+1$
$=3{x}^{4}+5{x}^{3}+4{x}^{2}+3x+1$ Therefore, the equation $\left[\left(3x+5\right)x+4\right]x+3x+1=3{x}^{4}+5{x}^{3}+4{x}^{2}+3x+1$ isan identifity. (b) Given: A fourth-degree polynomial in x such as $3{x}^{4}+5{x}^{3}+4{x}^{2}+3x+1$ contains all of the powers of xfrom the first through the fourth. However, any polynomial can be written without powers of x. Evaluating a polynomial without powers of x (Horner's method) is somewhat easier than evaluating a polynomial with powers. Formula used: Grouping of similar terms in a polynomial, Suppose, an equation is given in the form $a{x}^{2}+2ax+3$ can be grouped and written as, $a{x}^{2}+2ax+3=ax\left(x+2\right)+3=a\left(x+2\right)x+3.$ This is known as grouping of an equation. Calculation: The polynomial, $P\left(x\right)=6{x}^{5}—3{x}^{4}+9{x}^{3}+6{x}^{2}—8x+12,$ is grouped by the similar terms and written as follow,
$P\left(x\right)=6{x}^{5}—3{x}^{4}+9{x}^{3}+6{x}^{2}—8x+12=x\left(6{x}^{4}-3{x}^{3}+9{x}^{2}-8x\right)+12$
$=x\left(x\left(6{x}^{3}-3{x}^{2}+9x+6\right)-8\right)+12$
$=x\left[x\left(x\left(6{x}^{2}-3x+9\right)+6\right)-8\right]+12$ Furhter, $x\left[x\left(x\left(6{x}^{2}-3x+9\right)+6\right)-8\right]+12=\left[\left(\left(6x-3\right)x+9\right)x+6\right]x-8x+12$ Therefore, the polynomial $P\left(x\right)=6{x}^{5}—3{x}^{4}+9{x}^{3}+6{x}^{2}—8x+12$ without powers is $\left[\left(\left(6x-3\right)x+9\right)x+6\right]x-8x+12$

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