Question

How do I do this Polynomial question?? PLEASE HELP

Answer 1

Answer:

See below for proof.

Step-by-step 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:

$\begin{aligned}ak+c & = 0\\\implies ak & = -c\\\implies k & = -\dfrac{c}{a} \end{aligned}$

$\begin{aligned}a+ck & = b\\\implies ck & = b-a\\\implies k & = \dfrac{b-a}{c} \end{aligned}$

Substitute the first equation into the second to eliminate k:

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

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

Answer 2

Answer:

Step-by-step 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

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

            ab  = a² - c²

Hence, proved.