Question

3xSQUARE+7xcube+12x+28
factorization explain​

Answer 1

The polynomial

$7 {x}^{3} + 3 {x}^{2} + 12x + 28$

is not factorable over rational numbers.

We can factorise it by calculating it's roots over real and complex numbers.

$7 {x}^{3} + 3 {x}^{2} + 12x + 28 = 0$

${x}^{3} + \frac{3}{7} {x}^{2} + \frac{12}{7} x + 4 = 0$

$(x + \frac{1}{7} ) ^{3} + \frac{81}{49} (x + \frac{1}{7} ) + \frac{1290}{343} = 0$

Therefore, the Cubic equation has only one real root.

All the roots are

$\boxed{x = - \omega_{k} \sqrt[3]{ \frac{645}{343} + \sqrt{ (\frac{645}{343} ) ^{2} + ( \frac{27}{49}) ^{3} } } - \omega_{k}^{2}\sqrt[3]{ \frac{645}{343} - \sqrt{( \frac{645}{343} ) {}^{2} + ( \frac{27}{49} ) {}^{3} } }- \frac{1}{7} }$

$x _{1} \approx - 1.351$

$x _{2,3} \approx0.461 \pm1.657i$