Given Polynomial → 3x² – 4x – 7 [tex]\\[/tex] For finding zeroes of any polynomial, first, we need to factorize it. [tex]\\[/tex] For factorizing 3x² – 4x – 7, find 2 numbers such that: Their product is -21 Sum is -4 The 2 numbers which satisfy the above criteria are +3 and -7 [tex]\\[/tex] So, by splitting the middle term, 3x² – 4x – 7 = 0 ⇒ 3x² + 3x – 7x – 7 = 0 ⇒ 3x (x + 1) -7 (x + 1) = 0 ⇒ (3x – 7) (x + 1) = 0 [tex]\\[/tex] Zeros of the polynomial: 3x – 7 = 0 ⇒ x = 7/3 — α x + 1 = 0 ⇒ x = -1 — β [tex]\\[/tex] Finding Sum of Zeros: α + β [tex]\implies \dfrac{7}{3}-1\\\\[/tex] [tex]\implies \dfrac{7-3}{3}\\\\[/tex] [tex]\implies \boxed{\bf{\alpha + \beta =\dfrac{4}{3}}}[/tex] Reply
Answer:
4/3 is the sum of two zeroes
Given Polynomial → 3x² – 4x – 7
[tex]\\[/tex]
For finding zeroes of any polynomial, first, we need to factorize it.
[tex]\\[/tex]
For factorizing 3x² – 4x – 7, find 2 numbers such that:
The 2 numbers which satisfy the above criteria are +3 and -7
[tex]\\[/tex]
So, by splitting the middle term,
3x² – 4x – 7 = 0
⇒ 3x² + 3x – 7x – 7 = 0
⇒ 3x (x + 1) -7 (x + 1) = 0
⇒ (3x – 7) (x + 1) = 0
[tex]\\[/tex]
Zeros of the polynomial:
⇒ x = 7/3 — α
⇒ x = -1 — β
[tex]\\[/tex]
Finding Sum of Zeros:
α + β
[tex]\implies \dfrac{7}{3}-1\\\\[/tex]
[tex]\implies \dfrac{7-3}{3}\\\\[/tex]
[tex]\implies \boxed{\bf{\alpha + \beta =\dfrac{4}{3}}}[/tex]