Question

Factorise and Divide : 10xy (14y^2 + 43y - 21) ÷ 5x (7y-3)

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

Let's factorise first,

14 × ( - 21 ) = - 294

By the L.C.M of 294 we get

294 = 2 × 7 × 3 × 7

Now, we have to find whose sum can be + 43.
( Remember that one should be - ve sunce porduct was - 294. )

= 49 - 6 = 43

Hence,

$= 10xy(14 {y}^{2} + 43y - 21) \\ = 10xy(14 {y}^{2} + 49y - 6y - 21) \\ = 10xy(7y(2y + 7) - 3(2y + 7)) \\ = 10xy(7y - 3)(2y + 7)$

Now, lets divide!


$= \frac{10xy(7y - 3)(2y + 7)}{5x(7 - 3)} \\ = 2y(2y + 7)$

$\textsf{\red{ Hence factors are 2y(2y + 7) }}$

$\textbf{ Have great future ahead! }$

Answer 2

14 × ( - 21 ) = - 294

By the L.C.M of 294 we get

294 = 2 × 7 × 3 × 7

Now, we have to find whose sum can be + 43.

( Remember that one should be - ve sunce porduct was - 294. )

= 49 - 6 = 43

Hence,

\begin{gathered} = 10xy(14 {y}^{2} + 43y - 21) \\ = 10xy(14 {y}^{2} + 49y - 6y - 21) \\ = 10xy(7y(2y + 7) - 3(2y + 7)) \\ = 10xy(7y - 3)(2y + 7)\end{gathered}

=10xy(14y

2

+43y−21)

=10xy(14y

2

+49y−6y−21)

=10xy(7y(2y+7)−3(2y+7))

=10xy(7y−3)(2y+7)

{10xy(7y - 3)(2y + 7)}{5x(7 - 3)} \\ = 2y(2y + 7)\end{gathered}

=

5x(7−3)

10xy(7y−3)(2y+7)

=2y(2y+7)