# If the polynomialx^4-6x^3+16x^2-25x+10 is divided by x^2-2x+k the remainder is x+a find k and a.?

Nov 25, 2017

$k = 5$ and $a = - 5$

#### Explanation:

From ${x}^{2} - 2 x + k = 0$, ${x}^{2} = 2 x - k$

After replacing ${x}^{2}$ into $2 x - k$ in polynomial and equating remainder into $x + a$,

${\left({x}^{2}\right)}^{2} - 6 x \cdot \left(2 x - k\right) + 16 \cdot \left(2 x - k\right) - 25 x + 10 = x + a$

${\left(2 x - k\right)}^{2} - 12 {x}^{2} + 6 k x + 32 x - 16 k - 25 x + 10 = x + a$

${\left(2 x - k\right)}^{2} - 12 {x}^{2} + 6 k x + 7 x - 16 k + 10 = x + a$

$4 {x}^{2} - 4 x k + {k}^{2} - 12 {x}^{2} + \left(6 k + 7\right) \cdot x - 16 k + 10 = x + a$

$- 8 {x}^{2} - 4 x k + {k}^{2} + \left(6 k + 7\right) \cdot x - 16 k + 10 = x + a$

$- 8 \cdot \left(2 x - k\right) - 4 x k + {k}^{2} + \left(6 k + 7\right) \cdot x - 16 k + 10 = x + a$

$- 16 x + 8 k + \left(2 k + 7\right) \cdot x + {k}^{2} - 16 k + 10 = x + a$

$\left(2 k - 9\right) \cdot x + {k}^{2} - 8 k + 10 = x + a$

After equating coefficients,

$2 k - 9 = 1$ and $a = {k}^{2} - 8 k + 10$

Hence $k = 5$ and $a = - 5$

Nov 27, 2017

$k = 5 , \mathmr{and} , a = - 5.$

#### Explanation:

The polynomial $p \left(x\right) = {x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 25 x + 10 ,$ when

divided by $\left({x}^{2} - 2 x + k\right) ,$ leaves the remainder $r \left(x\right) = \left(x + a\right) .$

If the quotient polynomial is $q \left(x\right) ,$ then, knowing that,

$p \left(x\right) = \left({x}^{2} - 2 x + k\right) q \left(x\right) + r \left(x\right) ,$ we have,

$p \left(x\right) - r \left(x\right) = \left({x}^{2} - 2 x + k\right) q \left(x\right) .$

This means that, $p \left(x\right) - r \left(x\right)$ is divisible by $\left({x}^{2} - 2 x + k\right) .$

So, when $p \left(x\right) - q \left(x\right)$ is divided by $\left({x}^{2} - 2 x + k\right) ,$ the Remainder

must be $0.$

Now,

$p \left(x\right) - r \left(x\right) = \left({x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 25 x + 10\right) - \left(x + a\right) , i . e . ,$

$p \left(x\right) - r \left(x\right) = {x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 26 x + \left(10 - a\right) .$

By Long Division of $p \left(x\right) - r \left(x\right)$ by $\left({x}^{2} - 2 x + k\right) ,$ we get,

$p \left(x\right) - r \left(x\right) = {x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 26 x + \left(10 - a\right) ,$

$= \left({x}^{2} - 2 x + k\right) \left\{{x}^{2} - 4 x + \left(8 - k\right)\right\} + \left\{2 \left(k - 5\right) x + {k}^{2} - 8 k + 10 - a\right\} .$

$\therefore \text{The Remainder=} 0 \Rightarrow \left\{2 \left(k - 5\right) x + {k}^{2} - 8 k + 10 - a\right\} = 0.$

$\therefore \left(k - 5\right) = 0 , \mathmr{and} , {k}^{2} - 8 k + 10 - a = 0.$

$\therefore k = 5 , \mathmr{and} , a = {k}^{2} - 8 k + 10 = 25 - 40 + 10 = - 5.$

Enjoy Maths.!