When a polynomial f(x) is divided by x-3 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) By the Remainder Theorem,we have $f(3) = 7$ and $f(-6) = 22.$Since the divisor $(x-3)(x+6)$ is a quadratic polynomial,the remainder will be of the

Vedclass · Accepted Answer

$\frac{-5}{3} x+12$

Vedclass · Answer

(A) By the Remainder Theorem,we have $f(3) = 7$ and $f(-6) = 22.$Since the divisor $(x-3)(x+6)$ is a quadratic polynomial,the remainder will be of the form $ax + b.$Thus,$f(x) = (x-3)(x+6)Q(x) + (ax + b).$Substituting $x = 3$: $f(3) = 3a + b = 7$ (Equation $1$).Substituting $x = -6$: $f(-6) = -6a + b = 22$ (Equation $2$).Subtracting Equation $2$ from Equation $1$: $(3a + b) - (-6a + b) = 7 - 22.$$9a = -15 \implies a = -\frac{15}{9} = -\frac{5}{3}.$Substituting $a = -\frac{5}{3}$ into Equation $1$: $3(-\frac{5}{3}) + b = 7 \implies -5 + b = 7 \implies b = 12.$Therefore,the remainder is $-\frac{5}{3}x + 12.$

When a polynomial $f(x)$ is divided by $x-3$ and $x+6,$ the respective remainders are $7$ and $22.$ What is the remainder when $f(x)$ is divided by $(x-3)(x+6)?$

Similar Questions

Factorize: $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b+c-a)$

If $2x + \frac{1}{2x} = 2$,then what is the value of $\sqrt{2(\frac{1}{x})^4 + (\frac{1}{x})^5}$?

If $\frac{x}{y} + \frac{y}{x} = 6,$ then find the value of $\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}}.$

Resolve into factors: $4 x^{2}+12 x y+9 y^{2}-8 x-12 y$

Resolve into factors: $(a+b)^{2}-14 c(a+b)+49 c^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module