Question

Factorise 2x

3

-5 x

2

- 23 x -10​

Answer 1

Answer:

235-3x

Step-by-step explanation:

2x+3-5x+2-23×-10

230+5-3x

235-3x

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Answer 2

Question :

$\sf \small 2 {x}^{3} - 5 {x}^{2} - 23x - 10$

Solution :

$\sf \small 2 {x}^{3} + {x}^{2} - 6 {x}^{2} - 3x - 20x - 10$

$\sf \small {x}^{2} (2x + 1) - 3x(2x + 1 ) - 10(2x + 1)$

Taking (2x + 1) as common we get :

$\sf \small (2x + 1)[ {x}^{2} - 3x - 10]$

Again we need to factorise the x² - 3x - 10 to get the result by middle term factorisation

$\sf \small(2x + 1)[x + (5 - 2)x - 10]$

The process for getting it clear are that how 5 and 2 came and why we subtracted it are given below :

Conditions :

When the third term has a negative sign it means that factors to be subtracted in middle term factorisation

When the third term is positive it means that factors to be added in the middle term factorisation

Here 10 is negative so factors of 10 i.e. 5 and 2 are subtracted to get 3

$\sf \small(2x + 1)[x + 5x - 2x - 10]$

Now we need to factorise

$\sf \small(2x + 1)[x(x + 5) - 2(x + 5)]$

$\sf \small(2x + 1)(x + 5)(x - 2)$

$\boxed{ \sf \small \: \therefore \: the \: answer \: is \: \pink{(2x + 1)(x + 5)(x - 2)}}$

More Information :

• In Quadratic Factorization using Splitting of Middle Term which is x term is the sum of two factors and product equal to last term.

• To Factor the form :ax2 + bx + c. Factor : 6x2 + 19x + 10.

1. Find the product of 1st and last term( a x c).