# What's The Remainder When Divided by (x-1)(x-2)

If a polynomial is divided by $(x-1)$ then remainder is 5 and if divided by $(x-2)$ the remainder is 7. What will be the remainder is the polynomial is divided by $(x-1)(x-2)$ ?

As the degree is unknown so we can't write the polynomial with arbitrary coefficients. So we have to assume the polynomial as $f(x)$. Now we can write ... $$f(x)=(x-1)g(x)+5$$ $$f(x)=(x-2)h(x)+7$$ Where $g(x)$ & $h(x)$ are some polynomial of x. then I subtract these two equations, but can't go further. Am I going correct? Should I need to use differentiation?

(If you are using some theorem please provide a link so that I can learn that)

By remainder theorem, “When f(x) is divided by (x−1), the remainder is 5” can be translated to:-

(1) … $$f(1) = 5$$

Similarly, we have:-

(2) … $$f(2) = 7$$

When $$f(x)$$ is divided by $$(x-1)(x-2)$$, then basically we have:-

(3) … $$f(x) = (x-1)(x-2) \times$$Quotient + Remainder

Since the degree of the remainder must be one lower than that of the divisor, [$$= 2$$ from $$(x-1)(x-2)$$], the remainder can have degree = 1 (or lower) only. The remainder should then take the form $$ax + b$$, which is a general expression of degree 1 (or lower if a = 0) in x. Therefore, (3) becomes:-

(4) …$$f(x) = (x-1)(x-2)Q(x) + (ax + b)$$

(1) and (2) can be used to find the values of $$a$$ and $$b$$ from (4).

HINT:

$f(x)$ can be written as $$(x-1)(x-2)g(x)+Ax+B=(x-1)(x-2)g(x)+C(x-1)+D(x-2)$$ where $g(x)$ is a finite polynomial.

$f(2)=7\,\Rightarrow\,f = 7+(x\!-\!2)g.\$ $5 = f(1) = 7-g(1)\,\Rightarrow\, g(1) = 2\,\Rightarrow\, g = 2+(x\!-\!1)h$

Therefore $\ f\, =\, 7+(x\!-\!2)\underbrace{(2+(x\!-\!1)h}_{\large g})\, =\, 2x+3 + (x\!-\!2)(x\!-\!1)h$

hint...Write $$f(x) =(x-1)(x-2)q(x)+ax+b$$ since the remainder on division by a quadratic will be of degree of at most one less. Use the known results to find $a$ and $b$.

$(ax + b)/(x-1)$, the quotient is a and the remainder is $a+b$ $(ax+b)/(x-2)$, the quotient is a and the remainder is $7$

so we have $a+b = 5$ and $2a+b = 7$, equating the two we have $a= 2$ and $b = 3$ so the remainder is $2x+3$ if said polynomial is divided by $(x-1)(x-2)$

Answer $2x+3$