The Question which you have written 3x^2+25x+24 is not FACTORISABLE. But if you make some changes in the constant term, "24" and change it to "8" , It's exactly factorisable in terms of LHS.
=>We HAVE to change RHS to 3x^2+25x+8 to make it Factoisable/
Answer :
Step-by-step explanation :
=> It can be then taken common as (x+8) (3x+1).
Thus, This is factorisable in form of (x + Something) (3x + something)
Answers & Comments
The Question which you have written 3x^2+25x+24 is not FACTORISABLE. But if you make some changes in the constant term, "24" and change it to "8" , It's exactly factorisable in terms of LHS.
=>We HAVE to change RHS to 3x^2+25x+8 to make it Factoisable/
Answer :![(x+8) (3x+1) (x+8) (3x+1)](https://tex.z-dn.net/?f=%28x%2B8%29%20%283x%2B1%29)
Step-by-step explanation :![3x^{2} + 25x + 8\\ ></p><p></p><p><strong>By Middle Term Splitting, Multiplying the coefficient of the first term by the constant 3 • 8 = 24</strong></p><p></p><p><strong>So, We need to Find two factors of 24 whose sum equals the coefficient of the middle term, which is 25 .</strong></p><p></p><p><strong>(Something) + (Something) = 25 </strong></p><p><strong>So, Such two numbers are 1 and 24. </strong></p><p><strong>As, 1 + 24 = 25 and 1 • 24 = 24</strong></p><p><strong>Thus,</strong> <img src=](https://tex.z-dn.net/?f=3x%5E%7B2%7D%20%2B%2025x%20%2B%208%5C%5C)
=> It can be then taken common as (x+8) (3x+1).
Thus, This is factorisable in form of (x + Something) (3x + something)