Question

Find the values of m and n so that the polynomial a³ - ma² - 13a + n has (a - 1)and
(a + 3) as factors​

Answer 1

Solution

Given :-

• Polynomial, a³ - ma² - 13a + n = 0
• Zeroes of given polynomial ( a -1) & (a+3)

Find :-

• Value of m & n

Explanation,

we know,

If ( a +3) & ( a-1) be zeros of given equation , its meant that value of a satisfied of given polynomial .

Case(1).

• when , a = 1

keep in equation,

➠ (1)³ - m * 1² - 13 * 1 + n = 0

➠ 1 - m - 13 + n = 0

➠ m - n = - 12 -----------(1)

Case(2).

• when, a = -3

Keep in equation

➠ (-3)³ - m * (-3)² - 13 * (-3) + n = 0

➠ -27 - 9m + 39 + n = 0

➠ 9m - n = 12 ------------(2)

Subtract equ(1) & equ(2)

➠ m - 9m = -12 - 12

➠ -8m = - 24

➠ m = (-24)/(-8)

➠ m = 3

Keep Value in equ(1)

➠ 3 - n = - 12

➠ n = 3 + 12

➠ n = 15

Hence

• Value of n = 15
• Value of m = 3

_________________

Answer 2

$\large\underline\bold{Question:-}$

Find the values of m and n so that the polynomial a³ - ma² - 13a + n has (a - 1)and (a + 3) as factors.

$\large\underline\bold{Answer:-}$

$\small\underline\bold{Given:-}$

$\sf\ Polynomial:- a^3-ma^2-13a+n$

$\sf\ Zeroes\:of\: polynomial = (a-1)(a+3)$

$\small\underline\bold{Required\:Solution:-}$

$\small\underline\bold\pink{Case\:1:-}$

$\sf\ Firstly, \:if \:a-1\: is\: factor$

$\sf\ then,\: a=1$

★$\small\bold{Putting\:the\:value}$

$\sf\ P(1)=13-m(12)-13(1)+n=0$

$\sf\ = -12-m+n=0 --- (1)$

$\small\underline\bold\pink{Case\:2:-}$

$\sf\ If \:a+3\: is \:factor \:then\:$

$\sf\ a=-3$

★$\small\bold{Putting\:the\:value}$

$\sf\ P(-3)=(-3)^3-m(-3)^2-13(-3)+n$

$\sf\ = 12-9m+n=0 ---(2)$

$\large\underline\bold{ From\: (1)\: and\: (2)\: we \:get,}$

⟹$\sf\ -12-m+n=12-9m+n$

⟹$\sf\ -m+n+9m-n=-24$

⟹$\sf\ -8m=-24$

⟹$\sf\ m=3$

$\small\bold{Keep\:the\: values\:from\:eq\:1}$

⟹$\sf\ n=12+m \: [using(1)]$

⟹$\sf\ n=12+3$

⟹$\sf\ n=15$

$\sf\ Hence,$

$\dagger\small\underline\bold\purple{Value\:of\:m= 3}$

$\dagger\small\underline\bold\purple{value\:of\:n= 15}$

$\rule{200}3$