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Question: Factorise\[{\text{: }}2{x^2} - 7x - 15\] . (A) \(\left( {x - 5} \right)\left( {2x + 3} \right)\) ...

Factorise2x27x15{\text{: }}2{x^2} - 7x - 15 .
(A) (x5)(2x+3)\left( {x - 5} \right)\left( {2x + 3} \right)
(B) (x5)(2x+1)\left( {x - 5} \right)\left( {2x + 1} \right)
(C) (x6)(2x+3)\left( {x - 6} \right)\left( {2x + 3} \right)
(D) (x6)(2x+3)\left( {x - 6} \right)\left( {2x + 3} \right)

Explanation

Solution

In this question, we have to factorise the given equation. Factorise means we have to make the factors of given equations which can be done by any of the three methods middle term splitting, method of completing squares and discriminant method. This question can be solved by any of these methods and we are going to solve it by a middle term splitting method.

Complete answer:
The given question is to factorise the equation 2x27x15 ..................... (1)2{x^2} - 7x - 15{\text{ }}.....................{\text{ (1)}}
For solving this, we will use a middle term splitting method. Which is :
Firstly, we will multiply the coefficient of x2{x^2} and the constant term that is
\Rightarrow 2×15=302 \times 15 = 30
Now we will make the factors of 3030 and factors of 3030 are 2×3×52 \times 3 \times 5.
Now, we have to arrange these 33 factors to 22 factors by multiplication in such a way that if we multiply those 22 factors we get 3030 (which is the constant term) and when we add or subtract these 22 factors.
We get 77 which is the coefficient of xx. So we arrange 2×3×52 \times 3 \times 5 to 10×310 \times 3 because 1010 and 33, on multiplication gives 3030 and 1010 and 33. On subtracting gives 77
So equation (1) becomes 2x2(103)x152{x^2} - (10 - 3)x - 15
In the above equation, we have written
\Rightarrow 77 as (103)(10 - 3)
\therefore On multiplying xx by both terms ad we get
\Rightarrow 2x210x+3x152{x^2} - 10x + 3x - 15
Now we get common 2x2x from first two terms and 33 from last two terms, we get
\Rightarrow 2x(x5)+3(x5)2x(x - 5) + 3(x - 5)
Now, we get 22 terms and from these two, we get (x5)(x - 5) common out of these two
\Rightarrow (x5) (2x + 3)(x - 5){\text{ (2x + 3)}}
Which is option (A)
If means factorization of 2x27x152{x^2} - 7x - 15 is
\Rightarrow (x5)(2x+3)\left( {x - 5} \right)\left( {2x + 3} \right)
And (x5) & (2x+3)(x - 5){\text{ \& }}\left( {2x + 3} \right) on multiplication gives 2x27x152{x^2} - 7x - 15

So, the correct option is A.

Note: Factorisation of one degree equation or linear equation gives the value of 11 variable used in the equation. Factorisation of two degree equations gives two factors and factorization of three degree equations gives three factors. Factorisation of three degree equation can be evaluated by deducing three degree equation to the multiplication of one & two degree equation and on further solving two degree equation we finally get 33 factors of cubic equation.