Polynomial Division and Remainder Theorem - Class IX

For f(x) to be divisible by x^2+x+1, it must vanish at the roots of x^2+x+1=0, which are omega and omega^2. Substituting x=omega into P(x^3)+Q(x^3) gives P(1)+Q(1)=0, implying P(1)=-Q(1). This condition is satisfied if both P(x) and Q(x) contain the factor (x-1).

Using polynomial division or substitution, set x^2=2. Then 3(x^2)^2 + 5(x^2) + 9 becomes 3(2)^2 + 5(2) + 9 = 3(4) + 10 + 9 = 12 + 10 + 9 = 31.

Division algorithm stated that a polynomial $f(x)$ can written as
   $f(x) = g(x)q + r$     where $q$ and $r$ are unique integer and $0 <= r < g(x)$.
Here,
$g(x)=2x+3$ and $r(x)=x-1$
The power of the remainder is always less than the power of the divisor
Here, the degree of remainder is $1$ and the degree of divisor is $1$, which is not possible. Thus, $(x-1)$ cannot be the remainder of $p(x)$ when divided by $(2x+3)$

Dividend = (divisor $\times$ quotient) + remainder
= $(3x^2, -, 2x, +, 1), \times,  (x, +, 2),+, (2x, -, 5)$
= $3x^3, +, 4x^2, -, x,-, 3$

Here, $p(x) = a{x}^{3}+b{x}^{2}+cx+d$ and factor of $ax+b$ is
$ax+b = 0$
$x = -\frac ba$
$ p(-\frac ba) = a(-\frac ba)^{3}+b(-\frac ba)^{2}+c(-\frac ba)+d$
$p(-\frac ba) = -\frac {b^3}{a^2}+ \frac {b^3}{a^2} -\frac {bc}{a}+d $
$p(-\frac ba) =  -\frac {bc}{a}+d $
$p(-\frac ba) = \frac 1a (ad - bc) $
when $a{x}^{3}+b{x}^{2}+cx+d$ is divided by $ax+b$ then remainder is $p(-\frac ba) = \frac 1a (ad - bc) $
Option B is correct.

Since $\displaystyle f\left ( x \right )=2x^{3}+4x^{2}+2ax+b$ is exactly divisible
by $x\displaystyle ^{2}-1=\left ( x-1 \right )\left ( x+1 \right )$
$\displaystyle \therefore f\left ( 1 \right )=0$ and $\displaystyle f\left ( -1 \right )=0$
These give
    $2+4+2a+b=0$
or $2a+b+6=0$          .....(i)
and $-2+4-2a+b=0$
or $2a-b-2=0$        ....(ii)
Solving equations (i) and (ii) we get 
$a=-1, b=-4$

$x^2y\times\dfrac{x}{y}=x^3\Rightarrow $when $x^2$ is divided by$ \dfrac{1}{x}\ $gives $  x^3$

deg $p(x)=$ deg $g(x)+r(c)$

Divide $x^3-x^2+x-1$ by   $x-1$

Consider the polynomial $f(x)=x^3-3x^2-x+3$ and factorise it as follows:

Consider the polynomial $f(x)=x^6+x^4-x^2-1$ and factorise it as follows:

Given f(y)=$3y^{4}+5y^{3}-7y^{2}+2y+2$ and (g)=$ y^{2}-4y+2$

$x^2-x+1)\overline {x^4-3x^2+4x+5}$ ( $x^2+x-3$
                  $\underline {\underset {-}{}x^4\underset {-}{+}x^2              \underset{+}{-}x^3}$
                  $x^3-4x^2+4x+5$
                  $\underline {\underset {-}x^3\underset {+}{-}x^2\underset {-}{+}x}$
                  $-3x^2+3x+5$
                  $\underline {\underset {+}{-}3x^2\underset {-}{+}3x\underset {+}{-}3}$
                                  $8$
Hence, Quotient=$x^2+x-3$
Remainder=8.

option (A) and have deg $r(x)=0$,

$f(x)=q(x)g(x)+r(x)$

$\therefore f(x)= (x-2)(x-1-x^2)+3$

$\Rightarrow  f(x)= x(x-1-x^2)-2(x-1-x^2)+3$

$=x^2-x-x^3-2x+2+2x^2+3$

$=-x^3+3x^2-3x+5$

according to division algorithm $p(x)=q(x)g(x)+r(x)$

By remainder theorem,
$f(x)=q(x)g(x)+r(x)$
$\therefore 2x^5+3x^4+4x^3+4x^2+3x+2=(2x^2+x+1)(x^3+x^2+x+1)+r(x)$
$=2x^2(x^3+x^2+x+1)+x(x^3+x^2+x+1)+1(x^3+x^2+x+1)+r(x)$
$=2x^5+2x^4+2x^3+2x^2+x^4+x^3+x^2+x+x^3+x^2+x+1+r(x)$
$=2x^5+3x^4+4x^3+4x^2+2x+1+r(x)$
$r(x)=x+1$

degree of $p(x)$ and $q(x)$ are equal in (A),(C),(D)

$4x^2+3x-2)\overline {8x^4+14x^3-2x^2+7x-8}$ ( $2x^2+2x-1$
                          $\underline {\underset {-}{8}x^4\underset {-}{+}6x^3\underset {+}{-}4x^2}$
                                     $8x^3+2x^2+7x-8$
                                     $\underline {\underset {-}{8}x^3\underset {-}{+}6x^2\underset {+}{-}4x}$
                                             $-4x^2+11x-8$
                                             $\underline {\underset {+}{-}4x^2\underset {+}{-}3x\underset {-}{+}2}$
                                                          $14x-10$
We have to add $10-14x$ so that $8x^4+14x^3-2x^2+7x-8$ is completely divisible by $4x^2+3x-2$.

$x^3-3x+1)\overline {x^5-4x^3+x^2+3x+1}$($x^2-1$
                         $\underset {-}{x^5}\underset {+}{-}3x^3\underset {-}{+}x^2$
                         $\overline {-x^3+3x+1}$
                         $\underline {\underset {+}{-}x^3\underset {-}{+}3x\underset {+}{-}1}$
                                              $2$
Since remainder is non-zero.
Therfore,$x^3-3x+1$ is not a factor of $x^5-4x^3+x^2+3x+1$

$f(x)=$ is divided by another polynomial
$x^2-2x+k)\overline {x^4-6x^3+16x^2-25x+10}(x^2-4x+(8-k)$
                       $\underline {\underset {-}{x^4}\underset {+}{-2x^3}\underset{-}{+}kx^2}$
                       $-4x^3+(16-k)x^2-25x+10$
                       $\underline {\underset {-}{-4x^3}\underset {-}{+8x^2}                      \underset{+}{-}4kx}$
                       $(8-k)x^2+(4k-25)x+10$
                       $\underline {\underset {-}(8-k)x^2+\underset {-}(2k-16)x\underset{-}{+}(8k-k^2)}$
                       $(2k-9)x+(k^2-8k+10)$
But remainder is given $x+a$
$\therefore x+a=(2k-9)x+(k^2-8k+10)$
On equating coefficient, we get
$2k-9=1\Rightarrow k=5$
and $a=k^2-8k+10\Rightarrow a=25-40+10=-5$
Hence, $k=5,a=-5$

Since $2x+3$ is a factor of the polynomial $p\left(x\right)=2x^3+9x^2-x-b$
Therefore, by Factor theorem $p\left(-\dfrac32\right)=0$
$\Rightarrow 2\left(-\dfrac32\right)^3+9\left(-\dfrac32\right)^2-\left(-\dfrac32\right)-b=0$

$4x^2+3x-2)\overline {8x^4+14x^3-2x^2+8x-12}$ ( $2x^2+2x-1$
                            $\underline {\underset {-}{8}x^4\underset {-}{+}6x^3\underset {+}{-}4x^2}$
                            $8x^3+2x^2+8x-12$
                            $\underline {\underset {-}{8}x^3\underset {-}{+}6x^2\underset {+}{-}4x}$
                            $-4x^2+12x-12$
                            $\underline {\underset {+}{-}4x^2\underset {+}{-}3x\underset {-}{+}2}$
                                        $15x-14$

$\therefore$ The expression that must be subtracted is $15x-14$
and the expression that must be added is $-(15x-14)=-15x+14$

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right)$

Thus, $-3x-5$ must be subtracted to make it exactly divisble

$p(x)=x^4+4x-2x^2+x^3-10$ and $q(x)=x^2+3x-3x^2+4x+12$ and $r(x)=14$

Consider the polynomial $f(x)=x^2+4x-21$ and factorise it as follows:

When ${ x }^{ 4 }-{ 2x }^{ 3 }+3{ x }^{ 2 }-ax+b$ is divide by $ x-1,$ remainder is $5.$
So, substituting for $x$ is $1,$ in the above, we get

It is given that $f(x)$ leaves the remainder $0$ if divided by $(x+3)$ and $(x+4)$, which implies that $(x+3)$ and $(x+4)$ are factors of $f(x)$.

Using division algorithm method we get the value of $Q = x - 12$.

$x^2-2x-3=(x+1)(x-3)$

According to factor theorem

$2002 = 91 \times 22$

(a) If p(x) is not divisible by g(x), then $r(x)\neq 0 \therefore$ (a) is not true
(b) If p(x) is divisible by g(x), the $r(x)=0$ for all x i.e., r(x) is zero polynomial whose degree is not defined.
$\therefore $(b) is not true
(c) is clearly true [$\because$ division algorithm rule]
Since degree of $r(x)<$ degree of g(x) or $r(x)=0$, but $g(x)\neq 0$.
(d) $\therefore r(x)=g(x)$ is not true.

Let the remainder be ax^2+bx+c. Using f(3)=15 and the condition for (x-1)^2, we set up a system of equations. Solving for the remainder when divided by (x-3)(x-1)^2 yields a linear term coefficient p = -2.

$p{\left( x \right)} = 1 + x + {x}^{3} + {x}^{9} + {x}^{27} + {x}^{81} + {x}^{243}$

Here, $x + 2$ is a factor of $x^3 + 10x^2 + mx + n$
$x =-2$
$(-2)^3 + 10(-2)^2 + m(-2) + n =0$
$ -8 + 40 = 2m - n $
$2m -n = 32$                     .....(i)
Again, $x-1 $ is a factor of $x^3 + 10x^2 + mx + n$
$x =1$
$1+10+m+n=0$
$m + n =-11$                    .....(ii)
Adding (i) and (ii). we get,
$3m = 21$
$m=7$
By putting m in (i). we get,
$2(7) - n = 32 $
$ - n = 18 $
   $n = -18$
Option C is correct.

f(x)=$ 2x^3+ax^2-bx+3$ 
At x=2 
f(2)=15 
f(1)=0 
f(x)=$ 2x^3+ax^2-bx+3$ 
f(1)= 2+a-b+3=0 
a-b+5=0......A 
f(x)= $2x^3+ax^2-bx+3$ 
f(2)=$ 2(2^3)+a(2^2)-2b+3=15 $
4a-2b=-4 
Multiply A by 2 and subtract from above equation 
4a-2b=-4 
2a-2b+10=0 
2a-10=-4 
2a= 6 
a=3 
From A 
3-b+5=0 
8-b=0 
b=8 
So a=3 and b=8

$x^{3} + 5x^{2} + 10k$
$= (x^{2} + 2)(x + 5) + 10k - 2x - 10$
$\Rightarrow 10k - 2x - 10 = -2x$
$\Rightarrow 10k - 10 = 0$ or $k = 1$.

    by remainder theorem,

$x^{100}+2x^{99}+k$ is exactly divisible by $(x+1)$

Divide the polynomial by x^2+1 by substituting x^2 = -1. The terms become powers of (-1), simplifying to -x.

Let $f(x)=3x^3-2x^2-7x+6$

Let $f\left( x \right) =x^{ 3 }+4x^{ 2 }-7x+6$
As $f\left( x \right) $ is divided by $x-1$, substituting $x=1$ in $f\left( x \right) $ we get
$f\left( 1 \right) =1^{ 3 }+4\cdot1^{ 2 }-7\cdot1+6=4$
Hence, $4$ is the remainder.

Given: equation $4x^3-8x^2-x+5$