Question

If the polynomials 2x3+ kx2 + 3x – 5 and x3+ x2– 2x + k, leave the same
remainder when divided by (x – 2), find the value of k.

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

$p(x) = {2x}^{3} + {kx}^{2} + 3x - 5$

and

$q(x) = {x}^{3} + {x}^{2} - 2x + k$

Now, given that when p(x) and q(x) divided by x - 2, both leaves the same remainder.

We know,

Remainder Theorem states that if a polynomial p(x) of degree greater than or equals to 1 is divided by x - a, then remainder is p(a).

So, using Remainder Theorem, we have

$\rm \: p(2) = q(2)$

$\rm \: {2(2)}^{3} + {k(2)}^{2} + 3(2) - 5 = {(2)}^{3} + {(2)}^{2} - 2(2) + k$

$\rm \: 16 + 4k + 6 - 5 = 8 + 4 - 4 + k$

$\rm \: 17 + 4k = 8 + k$

$\rm \: 4k - k = 8 - 17$

$\rm \: 3k = - 9$

$\rm\implies \:k \: = \: - \: 3$

$\rule{190pt}{2pt}$

Additional Information :-

Factor Theorem :- This theorem states that if a polynomial f(x) of degree greater than or equals to 1 is divided by x - a, then f(a) = 0.

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$