Advertisements
Advertisements
प्रश्न
If 2 and -2 are two zeroes of the polynomial `(x^4 + x^3 – 34x^2 – 4x + 120)`, find all the zeroes of the given polynomial.
उत्तर
Let f(x) = x^4 + x^3 – 34x^2 – 4x + 120
Since 2 and -2 are the zeroes of f(x), it follows that each one of (x – 2) and (x + 2) is a factor of f(x).
Consequently, (x – 2) (x + 2) = (x2 – 4) is a factor of f(x).
On dividing f(x) by `(x^2 – 4)`, we get:
f(x) = 0
⇒` (x2 + x – 30) (x^2 – 4) = 0`
⇒` (x^2 + 6x – 5x – 30) (x – 2) (x + 2)`
⇒ `[x(x + 6) – 5(x + 6)] (x – 2) (x + 2)`
⇒`(x – 5) (x + 6) (x – 2) (x + 2) = 0`
⇒` x = 5 or x = -6 or x = 2 or x = -2`
Hence, all the zeroes are 2, -2, 5 and -6.
APPEARS IN
संबंधित प्रश्न
If α and β are the zeros of the quadratic polynomial p(y) = 5y2 − 7y + 1, find the value of `1/alpha+1/beta`
Find the quadratic polynomial, sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
If 𝛼, 𝛽 are the zeroes of the polynomial f(x) = x2 + x – 2, then `(∝/β-∝/β)`
Zeroes of a polynomial can be determined graphically. No. of zeroes of a polynomial is equal to no. of points where the graph of polynomial ______.
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the product of the other two zeroes is ______.
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
`2x^2 + (7/2)x + 3/4`
For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
`(-8)/3, 4/3`
Given that `sqrt(2)` is a zero of the cubic polynomial `6x^3 + sqrt(2)x^2 - 10x - 4sqrt(2)`, find its other two zeroes.
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
`y^2 + 3/2 sqrt(5)y - 5`
Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients.
