Math, asked by dony1301, 1 year ago

if ax3+bx2+x-6 has x+2 as factor and leaves the remainder 4 when divided by (x-2), find the values of 'a' and 'b' .

Answers

Answered by abhi178
565
Let p(x) = ax³ + bx² + x - 6

A/C to question,
(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0
So, p(2) = a(-2)³ + b(-2)² - 2 - 6 = 0
⇒-8a + 4b - 8 = 0
⇒ 2a - b + 2 = 0 -------------(1)

again, question said that if we p(x) is divided by ( x -2) then it leaves remainder 4.
so, P(2) = a(2)³ + b(2)² + 2 - 6 = 4
⇒8a + 4b - 4 = 4
2a + b -2 = 0 -------------(2)

solve equations (1) and (2),
4a = 0 ⇒a = 0 and b = 2

Then, equation will be 2x² + x - 6

Nishtha02: its not correct in (ii) as 8a+4b-4=4 so 2a+b-1=0
abhi178: 8a + 4b - 4 = 4 => 8a +4b -4 - 4 = 0 => 4(2a + b -2) = 0 => 2a + b - 2 = 0
abhi178: Well you can check the answer too okay . It's correct . see sincerely
Nishtha02: k thanx bro.... u r right
Answered by MARTINNN
265

Solution :

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