Is (x-3) a factor of f(x)=x^4-4x^3-x^2+16x-12
Answers
Answer:
Appropriate Question :-
If α and β are the zeroes of the quadratic polynomial f(x) = x² - px + q, then find the value of α² + β².
Given :-
α and β are the zeroes of the quadratic polynomial f(x) = x² - px + q.
To Find :-
What is the value of α² + β².
Solution :-
Given equation :
\dashrightarrow \sf\bold{x^2 - px + q}⇢x
2
−px+q
where,
a = 1
b = - p
c = q
Now, we have to find the sum and product of the zeroes :
\clubsuit♣ Sum of Zeroes :
\begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Sum\: of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}\\\end{gathered}
⟼
SumofZeroes(α+β)=
a
−b
Then,
\implies \sf \alpha + \beta =\: \dfrac{- (- p)}{1}⟹α+β=
1
−(−p)
\begin{gathered}\implies \sf \alpha + \beta =\: \dfrac{p}{1}\: \bigg\lgroup \bold{\purple{- \times - =\: +}} \bigg\rgroup\\\end{gathered}
⟹α+β=
1
p
⎩
⎪
⎪
⎪
⎧
−×−=+
⎭
⎪
⎪
⎪
⎫
\implies \sf\bold{\green{\alpha + \beta =\: p}}⟹α+β=p
Again,
\clubsuit♣ Product of Zeroes :
\begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Product\: of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}\\\end{gathered}
⟼
ProductofZeroes(αβ)=
a
c
Then,
\implies \sf \alpha\beta =\: \dfrac{q}{1}⟹αβ=
1
q
\implies \sf\bold{\green{\alpha\beta =\: q}}⟹αβ=q
Now, we have to find the value of α² + β² :
As we know that :
\longmapsto \sf\boxed{\bold{\pink{a^2 + b^2 =\: (a + b)^2 - 2ab}}}⟼
a
2
+b
2
=(a+b)
2
−2ab
We have :
α + β = p
αβ = q
According to the question by using the formula we get,
\leadsto \sf \alpha^2 + \beta^2 =\: (\alpha + \beta)^2 - 2\alpha\beta⇝α
2
+β
2
=(α+β)
2
−2αβ
\leadsto \sf \alpha^2 + \beta^2 =\: (p)^2 - 2q⇝α
2
+β
2
=(p)
2
−2q
\leadsto \sf\bold{\red{\alpha^2 + \beta^2 =\: p^2 - 2q}}⇝α
2
+β
2
=p
2
−2q
\therefore∴ The value of α² + β² is p² - 2q.
Answer:
Step by Step Solution
STEP1:
Equation at the end of step 1
((((x4) - 22x3) - x2) + 16x) - 12
STEP2:
Polynomial Roots Calculator :
2.1 Find roots (zeroes) of : F(x) = x4-4x3-x2+16x-12
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0