Question

Find m and n if x- 1 and x-2
exactly divide the polynomial x^3+mx^2-nx+10​

Answer 1

EXPLANATION.

x - 1 and x - 2 are the zeroes of the polynomial.

Polynomial x³ + mx² - nx + 10.

As we know that,

x - 1 is the zeroes of the polynomial, we get.

⇒ x - 1 = 0.

⇒ x = 1.

Put the value of x = 1 in equation, we get.

⇒ x³ + mx² - nx + 10 = 0.

⇒ (1)³ + m(1)² - n(1) + 10 = 0.

⇒ 1 + m - n + 10 = 0.

⇒ m - n + 11 = 0.

⇒ m = n - 11. ⇒ (1).

x - 2 is the zeroes of the polynomial, we get.

⇒ x - 2 = 0.

⇒ x = 2.

Put the value of x = 2 in equation, we get.

⇒ x³ + mx² - nx + 10 = 0.

⇒ (2)³ + m(2)² - n(2) + 10 = 0.

⇒ 8 + 4m - 2n + 10 = 0.

⇒ 4m - 2n + 18 = 0.

⇒ 2(2m - n + 9) = 0.

⇒ 2m - n + 9 = 0. ⇒ (2).

From equation (1) & (2) we get,

Put the value of equation (1) in equation (2), we get.

⇒ 2(n - 11) - n + 9 = 0.

⇒ 2n - 22 - n + 9 = 0.

⇒ n - 13 = 0.

⇒ n = 13.

Put the value of n = 13 in equation (1), we get.

⇒ m = n - 11.

⇒ m = 13 - 11.

⇒ m = 2.

Value of m = 2 & n = 13.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \ \: :  ⟼ Let \: f(x) = {x}^{3} + {mx}^{2} - nx + 10$

$\large\underline{\bold{❥︎Step :- 1 }}$

$\begin{gathered}\tt\red{According \: to \: statement}\end{gathered}$

$\begin{gathered}\red{\tt \ \: :  ⟼ (x \: - 1 )\: divides \: f(x)}\end{gathered}$

$\tt\implies \:f(1) = 0$

$\tt\implies \: {(1)}^{3} + m {(1)}^{2} - n \times 1 + 10 = 0$

$\tt \ \: :  ⟼ 1 + m - n + 10 = 0$

$\tt \ \: :  ⟼ n \: - \: m \: = 11$

$\tt\implies \:n \: = 11 + m - - - (1)$

$\large\underline{\bold{❥︎Step :- 2}}$

$\begin{gathered}\tt\green{According \: to \: statement}\end{gathered}$

$\begin{gathered}\bf\green{\tt \ \: :  ⟼ (x - 2) \: divides \: f(x)}\end{gathered}$

$\tt\implies \:f(2) = 0$

$\tt \ \: :  ⟼ {(2)}^{3} + m {(2)}^{2} - 2 \times n + 10 = 0$

$\tt \ \: :  ⟼ 8 \: + {4m} - 2n + 10 = 0$

$\tt \ \: :  ⟼ 4m - 2n + 18 = 0$

$\tt\implies \:2m - n + 9 = 0$

$\tt \ \: :  ⟼ 2m - (11 + m) + 9 = 0 \: (using \: (1))$

$\tt \ \: :  ⟼ 2m - 11 - m + 9 = 0$

$\tt \ \: :  ⟼ m - 2 = 0$

$\bf\implies \:m = 2$

☆ On substituting the value of m in equation (1), we get

$\tt \ \: :  ⟼ n = 11 + 2 = 13$

$\begin{gathered}\begin{gathered}\bf Hence \: - \begin{cases} &\tt{m = 2} \\ &\tt{n = 13} \end{cases}\end{gathered}\end{gathered}$