# How do you solve 3x^4+11x^3=4x^2?

Mar 27, 2016

$\therefore x = 0 , 0 , \frac{1}{3} , - 4$

#### Explanation:

${x}^{2} \left(3 {x}^{2} + 11 x - 4\right) = 0$
$= {x}^{2} \left(3 {x}^{2} + 12 x - x - 4\right) = 0$
=x^2(3x(x+4)-1(x+4)=0
$= {x}^{2} \left(3 x - 1\right) \left(x + 4\right) = 0$
$\therefore x = 0 , 0 , \frac{1}{3} , - 4$

Mar 27, 2016

$x = 0$ or $x = \frac{1}{3}$ or $x = - 4$

#### Explanation:

We have $3 {x}^{4} + 11 {x}^{3} = 4 {x}^{2}$ and now let us bring the monomial to LHS, which gives us

$3 {x}^{4} + 11 {x}^{3} - 4 {x}^{2} = 0$ and now taking #x^2 common this becomes

${x}^{2} \left(3 {x}^{2} + 11 x - 4\right) = 0$

Now let us factorize $3 {x}^{2} + 11 x - 4$, for which we have to identify two numbes whose sum is $+ 11$ and product is $3 \times \left(- 4\right) = - 12$. Clearly these are $+ 12$ and $- 1$ and splitting middle term, we get

${x}^{2} \left(3 {x}^{2} + 12 x - x - 4\right) = 0$ or

${x}^{2} \left(3 x \left(x + 4\right) - 1 \cdot \left(x + 4\right)\right) = 0$ or

${x}^{2} \left(3 x - 1\right) \left(x + 4\right) = 0$

Hence, either $x = 0$ or $3 x - 1 = 0$ or $x + 4 = 0$

i.e. $x = 0$ or $x = \frac{1}{3}$ or $x = - 4$