Question

if f(x)=ax^2+bx+a is divided by (x-b) where a,b does not equal to 0,then a=?​

Answer 1

Question:-

If f(x) = ax² + bx + a is divided by (x - b) a , b ≠ 0 , remainder is a , then a is :

A) b²

B) - 1

C) 0

D) a + b

Answer:-

Given:

If f(X) = ax² + bx + a is divided by (x - b) leaves the remainder a where a , b ≠ 0.

We are given,

f(x) = ax² + bx + a

g(x) = x - b

g(x) = 0

⟹ x - b = 0

⟹ x = b

Substitute x = b in the given polynomial.

⟹ f(b) = a(b)² + (b)(b) + a

⟹ a = ab² + b² + a

⟹ a - a = ab² + b²

⟹ 0 = b² (a + 1)

⟹ 0 × 1/b² = a + 1

⟹ 0 = a + 1

⟹ a = - 1

∴ The required value of a is - 1 (Option - B).

Answer 2

Quêstioñ

if f(x)=ax²+bx+a is divided by (x-b) a,b≠0,

remainder is 'a' then

(a)b²

(b)-1

(c)0

(d)a+b

Given a quadratic equation is ax²+bx+a=0

$\sf \: let \: two \: roots \: be \: \alpha \: and \beta \\ \sf \longmapsto \: \alpha = - \beta \\ \sf \longmapsto \: \alpha + \beta - \frac{ - b}{a} \\ \sf \longmapsto \: - \beta + \beta = \frac{ - - b}{a} \\ \\ \tt \sf \longmapsto \: 0 \longmapsto \frac{ - b}{a} \\ \tt \sf \longmapsto \: b = 0 \: \: \: \: \: - - - - (1) \\ \tt \sf \: we \: have \: one \: root \: negative \: so \: \\ \sf \tt \longmapsto \alpha \beta <0 \\ \tt \sf \longmapsto \frac{c}{a}<0 \\ \\ \tt \sf \: so here \: either \: c \: or \: a \: will \: be \: negative \: \\ \tt \sf \: means \: \: a \: and \: c \: will \: be \: have \: oppositive \: sign \:$

f(b)=a(b)²+(b)(b)+a

a=ab²+b²+a

a-a=ab²+b²

0=b²(a+1)

0=a+1

a=-1