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How do you factor $2{x^2} + 9x + 4?$

Answer
VerifiedVerified
490.8k+ views
Hint:We can find the factors of a quadratic equation $\left( {a{x^2} + bx + c = 0} \right)$ by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The quadratic equations have two factors. we can find the factors either by the above formula or by factoring method where we split the middle term and by taking common, we can find the respective factors.

Complete step by step solution:
We are given a quadratic equation and equating the equation to zero $2{x^2} + 9x + 4 = 0$
Finding factors by splitting the middle term, in this method we have to find two numbers whose addition is the coefficient of x here the addition must be $ + 9$ and multiplication is the constant part multiplied with the coefficient of ${x^2}$ here the multiplication must be $4 \times 2 = 8$. Now, these two numbers are $8$ and $1$ which satisfies the above two conditions
Hence, we are splitting the middle term
$2{x^2} + 8x + x + 4 = 0$
Now, taking common from the first two terms and the last two terms respectively,
$2x\left( {x + 4} \right) + 1\left( {x + 4} \right) = 0$
Now, $\left( {2x + 1} \right)\left( {x + 4} \right) = 0$
The factors of the above given equation are
$
x + 4 = 0 \\
\Rightarrow x = - 4 \\
$
And
$
2x + 1 = 0 \\
\Rightarrow x = - \dfrac{1}{2} \\
$
Hence, factors are $ - 4$ and $ - \dfrac{1}{2}$

Note: You have to be very careful while selecting the two numbers. Signs should also be taken care of while selecting the numbers. The sum of the selected numbers is the coefficient of x and the product of numbers is the constant term and the coefficient of ${x^2}$. If we are unable to find the numbers use the formula above written in the hint part.