Question

Find the zeroes of cubic polynomials
(i) -x³ (ii) x² - x³ (iii) x³ - 5x² + 6x
without drawing the graph of the polynomial​

Answer 1

✰ Qᴜēsᴛíõñ :-

Find the zeroes of cubic polynomials

(i) -x³ (ii) x² - x³ (iii) x³ - 5x² + 6x

without drawing the graph of the polynomial

✪ Söʟúᴛîøɴ :-

(i) Given polynomial is y = -x³

➙ f(x) = -x³; f(x) = 0

➙ x³ = 0

➙ x = 3√0

➙ x = 0

⛬ Zero of the polynomial f(x) is only one i.e., 0

(ii) Given that y = x² - x³

➙ f(x) = x² (1 - x)

➙ f(x) = 0

➙ x² (1 - x) = 0

➙ x² = 0 and 1 - x = 0

➙ x = 0 and x = 1

⛬ Zero of the polynomial f(x) are two i.e., 0 and 1

(iii) Given that x³ - 5x² + 6x

Let f(x) = x³ - 5x² + 6x

➙ x(x² - 5x + 6)

➙ x(x² - 2x - 3x + 6)

➙ x[x(x - 2) - 3(x - 2)]

➙ x(x - 2) (x - 3)

⛬ Zero of the polynomial f(x) are x = 0, x = 2 and x = 3

Answer 2

given,

$1) { - x}^{3}$

$f(x) = { - x}^{3}$

${x}^{ 3} = 0(f(x) = 0)$

$x = 3 \sqrt{0}$

$x = 0$

-----------------------------------------------

$2)y = {x}^{2} - {x}^{3}$

$f(x) = {x}^{2} (1 - x)$

$f(x) = 0$

${x}^{2} (1 - x) = 0$

${x}^{2} = 0$

$1 - x = 0$

$x = 0 \: and \: x = 1$

---------------------------------------------------------

$3)y = {x}^{3} - 5 {x}^{2} + 6x$

$f(x) = {x}^{3} - 5 {x}^{2} + 6x$

$= x( {x}^{2} - 5x + 6)$

$= x( {x}^{2} - 2x - 3x + 6)$

$= x(x - x - 2) - 3(x - 2)$

$= x(x - 2)(x - 3)$

so,

$x = 0$

$x = 2 \: and \: x = 3$