Find the value of [tex] x^3 -ax^2 +2a^2x+4a^3 [/tex] when [tex] \dfrac{x}{a} = 1-\sqrt{-3} [/tex]
Home
/
tex]
Find the value of [tex] x^3 -ax^2 +2a^2x+4a^3 [/tex] when [tex] \dfrac{x}{a} = 1-\sqrt{-3} [/tex]
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Step-by-step explanation:
♦ Given ♦ :-
♦ To find ♦ :-
♦ Solution ♦ :-
Given polynomial is x³-ax²+2a²x+4a³
Let P(x) = x³-ax²+2a²x+4a³
Given that x/a = 1-√-3
=> x = a(1-√-3)
On cubing both sides then
=> x³ = [ a(1-√-3)]³
=> x³ = a³(1-3√-3-3√-3+3(-3))
=> x³ = a³(1-6√-3-9)
=> x³ = a³(-6√-3-8)
and
x² = [a(1-√-3)]²
=> x² = a²(1-3+2√-3)
=> x² = a²(-2+2√-3)
Now the value of P(x) at x = a(1-√-3) then
[a(1-√-3)]³-a[a(1-√-3)]²+2a²[a(1-√-3)]+4a³
= a³(-6√-3-8)-a[a²(-2+2√-3)]+2a²[a(1-√-3)]+4a³
= -6√-3 a³-8a³+2a³-2√-3 a³+2a³-2√-3 a³+4a³
= (-8a³+2a³+2a³+4a³)+(-6√-3 a³-2√-3 a³-2√-3 a³)
= (-8a³+8a³)+(-10√-3 a³)
= 0 + (-10√-3 a³)
= -10√-3 a³
♦ Answer ♦ :-
The required value of the polynomial is (-10√-3) a³
♦ Used Formulae ♦ :-
Verified answer
[tex]x^{3}-ax^{2}+2a^{2}x+4a^{3}=\boxed{0}[/tex]
[tex]\;[/tex]
Explanation
The Equation of Division
[tex]\boxed{P(x)=D(x)Q(x)+R(x)}[/tex]
[tex]\;[/tex]
Solution
Let [tex]k=x^{3}-ax^{2}+2a^{2}x+4a^{3}[/tex]
Then [tex]\dfrac{k}{a^{3}}=\bigg(\dfrac{x}{a}\bigg)^{3}-\bigg(\dfrac{x}{a}\bigg)^{2}+2\bigg(\dfrac{x}{a}\bigg)+4[/tex]
[tex]\;[/tex]
Let's substitute [tex]t=\dfrac{x}{a}[/tex]
[tex]\Longrightarrow \dfrac{k}{a}=t^{3}-t^{2}+2t+4[/tex]
[tex]\;[/tex]
We can square both sides of [tex]t=1-\sqrt{3}i[/tex]
That is, [tex]t-1=-\sqrt{3}i[/tex]
[tex]\Longrightarrow (t-1)^{2}=(-\sqrt{3}i)^{2}[/tex]
[tex]\Longrightarrow t^{2}-2t+1=-3[/tex]
[tex]\Longrightarrow\text{$t$ satisfies }t^{2}-2t+4=0[/tex]
[tex]\;[/tex]
We want to know [tex]t^{3}-t^{2}+2t+4=(t^{2}-2t+4)Q(t)+R(t)[/tex]
By actual division, [tex]t^{3}-t^{2}+2t+4=(t^{2}-2t+4)(t+1)[/tex]
As [tex]t^{2}-2t+4=0[/tex] we obtain [tex]t^{3}-t^2+2t+4=0[/tex]
[tex]\;[/tex]
Recall that [tex]t[/tex] is a substitution of [tex]\dfrac{k}{a}=t^{3}-t^{2}+2t+4[/tex]
[tex]\text{$\Longrightarrow \dfrac{k}{a^{3}}=0$}[/tex]
[tex]\text{$\Longrightarrow k=0$}[/tex]
[tex]\;[/tex]
Recall that [tex]k[/tex] is a substitution of [tex]k=x^{3}-ax^{2}+2a^{2}x+4a^{3}[/tex]
[tex]\Longrightarrow x^{3}-ax^{2}+2a^{2}x+4a^{3}=\boxed{0}[/tex]
[This is the required answer.]