Question

How do I do this Polynomial question??
PLEASE HELP

Answer 1

Explanation:

Polynomial:

p(x) = ax³ + bx + x

Let g(x) = x² + kx + 1 .

g(x) is a factor of p(x). So, p(x) is divided by g(x), the remainder will be 0.

Divide p(x) by g(x) using long division method. {attached as an image}.

By doing long division, we get the remainder.

Remainder = bx - ax + k²ax + ka + c

= (b - a + k²a)x+ [ka + c]

Remainder = 0

b - a + k²a = 0 and ka + c = 0

ka + c = 0

ka = -c

k = -c/a ----------------(I)

b - a + k²a = 0

`\sf b -a + (c^2)/(a^2)*a=0 -------[From \ (I)]\\\\b - a + (c^2)/(a)=0\\\\Multiply \ the \ above \ equation\ by \ a \\\\ab - a^2 + c^2 = 0\\\\\\`

ab = a² - c²

Hence, proved.

Answer 2

See below for proof.

Explanation:

Given polynomial:

`P(x)=ax^3+bx+c`

`\textsf{If }(x^2+kx+1)\:\textsf {is a factor of }P(x)\:\textsf {then}:`

`P(x)=(ax+c)(x^2+kx+1)`

Expand the brackets:

`\begin{aligned}\implies P(x) & = (ax+c)(x^2+kx+1)\\& = ax(x^2+kx+1)+c(x^2+kx+1)\\& = ax^3+akx^2+ax+cx^2+ckx+c\\& = ax^3+(ak+c)x^2+(a+ck)x+c\end{aligned}`

Compare coefficients of the x² and x terms in the expanded function with those of the original function:

`x^2: \quad ak+c=0`

`x: \quad a+ck=b`

Rewrite both equations to make k the subject:

`\begin{aligned}ak+c & = 0\\\implies ak & = -c\\\implies k & = -(c)/(a) \end{aligned}`

`\begin{aligned}a+ck & = b\\\implies ck & = b-a\\\implies k & = (b-a)/(c) \end{aligned}`

Substitute the first equation into the second to eliminate k:

`\begin{aligned}\implies -(c)/(a) & =(b-a)/(c)\\-c^2 & = a(b-a)\\-c^2 &=ab-a^2\\a^2-c^2 &=ab\end{aligned}`

Hence proving that
`a^2-c^2=ab`.