Polynomials

If α and β are zeroes of x² - 5x + k, and α - β = 1, find k. [CBSE 2025]

Find zeroes of 4x² - 4x + 1. [CBSE 2024]

If one zero of 3x² + 8x + k is 2/3, find other. [CBSE 2023]

If α and β are zeroes of polynomial x² - 5x + k, and α - β = 1, find value of k. [CBSE 2025]

Find zeroes of polynomial 4x² - 4x + 1 and verify relationship between zeroes and coefficients. [CBSE 2024]

If one zero of polynomial 3x² + 8x + k is 2/3, find the other zero. [CBSE 2023]

Divide 2x⁴ - 3x³ - 3x² + 6x - 2 by x² - 2. [CBSE 2022]

Find a quadratic polynomial whose zeroes are 2 and -3. [CBSE 2021]

α + β = -b/a

αβ = c/a

General form: ax² + bx + c, where a ≠ 0

If α and β are zeroes of ax² + bx + c:\nSum of zeroes: α + β = -b/a\nProduct of zeroes: αβ = c/a

For ax³ + bx² + cx + d with zeroes α, β, γ:\nα + β + γ = -b/a\nαβ + βγ + γα = c/a\nαβγ = -d/a

If p(x) and g(x) are polynomials with g(x) ≠ 0:\np(x) = g(x)×q(x) + r(x)\nwhere degree of r(x) < degree of g(x)

If α and β are zeroes of a quadratic polynomial, then the polynomial is: k[x² - (α+β)x + αβ] where k is any non-zero constant. This comes from (x-α)(x-β) = x² - (α+β)x + αβ.

When polynomial p(x) is divided by (x-a), the remainder is p(a). This can be proved using division algorithm: p(x) = (x-a)q(x) + r. Substituting x=a gives p(a) = 0 + r, so r = p(a).

If α and β are zeroes, then polynomial is: k[x² - (α+β)x + αβ] where k is any non-zero constant

If polynomial p(x) is divided by (x-a), remainder = p(a)