Question

factorise the question. 2x²+9x+10​

Answer 1

To Factorize:2x²+9x+10

=> {Firstly,to factorize a "quad eqn",you need to multiply the 1st and the 3rd term.Here,the number we obtain is 20. By factorizing it(using the prime division method),we have to find such a number that,while adding,gives us the number of the 2nd term. Here,we can use 5 and 4. By adding them,it's the second term,i.e,9. {9x} And when we multiply 5 and 4,we get 20.[As the product of the first and the last term is 20,in our quad eqn.] So,our factors are correct. }

Solution:-->

2x²+9x+10=0

(ax²+bx+c)=0 (A quadratic equation is always in this form.)

2x²+4x+5x+10=0 (Remember to take out the common terms from both the sides.

2x(x+2)+5(x+2)=0  (Make sure that the factors on the right side{x+2}are the same.)

(2x+5)(x+2)=0

when 2x+5=0

x=-5/2.

when x+2=0

x=-2.

So,the factors are:---> -5/2 and -2.

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Answer 2

$\huge\bold{\mathtt{\purple❥{ A{\pink{ N{\green{ S{\blue{ W{\red{ E{\orange{ R}}}}}}}}}}}}}$

$\huge{\sf{\pink{Explanation-}}}$

$\longrightarrow$ Firstly to factorise the "quad eqn" , you need to multiply the 1st and the 3rd term. Here the number we obtain is 20. By factorising it (using the prime division method), we have to find such a number that, while adding, gives us the number of second term. Here we can use 5 and 4. By adding them, it's the second term, i.e, 9. [9x]. And when we multiple 5 and 4, we get 20. (As the product of the first and last term is 20, in our quad eqn). So, our factors are correct.

$\huge\bold\red{SolutiOn-}$

${2x² + 9x + 19 = 0}$

${(ax² + bx + c)=0}$ [ A quadratic equation is always in this form ]

${2x²+4x+5x+10=0}$ [Remember to take our the common terms from the both sides ]

${2x(x+2)+5(x+2)=0}$ [ Make sure that the factors on the right side {x+2} are the same ]

${(2x+5)(x+2)=0}$

When, ${2x+5=0}$

${x=\frac{-5}{2}}$

When, ${x+2=0}$

. ° . ${x=0}$

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