Math Problems

MathematicsPrecalculusCubic Equations → Finding Factors of Basic Polynomials

Factor the polynomial

## Questions

 $$y = 64 c^{3} + 27$$ $$y = 27 b^{3} + 1$$ $$y = - 27 b^{12} + c^{6}$$ $$y = 8 y^{9} - 1$$ $$y = 64 c^{6} - 27 x^{12}$$ $$y = c^{9} - 1$$ $$y = a^{3} + y^{6}$$ $$y = 27 x^{6} + 8$$ $$y = 64 a^{6} - x^{12}$$ $$y = x^{4} - 4$$ $$y = 4 a^{6} - 1$$ $$y = b^{4} - 9$$ $$y = 27 b^{6} + 8$$ $$y = 4 x^{6} - 9 y^{12}$$ $$y = 27 a^{9} + 64$$ $$y = - 8 a^{6} + b^{3}$$ $$y = 27 b^{6} + 64$$ $$y = b^{4} - 1$$ $$y = c^{6} - 8$$ $$y = - a^{8} + 16 c^{4}$$ $$y = 27 b^{9} + x^{18}$$ $$y = 4 a^{4} - 9 x^{8}$$ $$y = 9 b^{4} - 4$$ $$y = 8 b^{3} - 27$$ $$y = 64 c^{18} + 27 y^{9}$$ $$y = - 16 b^{12} + 9 c^{6}$$ $$y = c^{12} + 8 y^{6}$$ $$y = - 27 c^{12} + x^{6}$$

 $$y = 64 c^{3} + 27$$ ⇒ $$\left(4 c + 3\right) \left(16 c^{2} - 12 c + 9\right)$$ $$y = 27 b^{3} + 1$$ ⇒ $$\left(3 b + 1\right) \left(9 b^{2} - 3 b + 1\right)$$ $$y = - 27 b^{12} + c^{6}$$ ⇒ $$\left(- 3 b^{4} + c^{2}\right) \left(9 b^{8} + 3 b^{4} c^{2} + c^{4}\right)$$ $$y = 8 y^{9} - 1$$ ⇒ $$\left(2 y^{3} - 1\right) \left(4 y^{6} + 2 y^{3} + 1\right)$$ $$y = 64 c^{6} - 27 x^{12}$$ ⇒ $$\left(4 c^{2} - 3 x^{4}\right) \left(16 c^{4} + 12 c^{2} x^{4} + 9 x^{8}\right)$$ $$y = c^{9} - 1$$ ⇒ $$\left(c^{3} - 1\right) \left(c^{6} + c^{3} + 1\right)$$ $$y = a^{3} + y^{6}$$ ⇒ $$\left(a + y^{2}\right) \left(a^{2} - a y^{2} + y^{4}\right)$$ $$y = 27 x^{6} + 8$$ ⇒ $$\left(3 x^{2} + 2\right) \left(9 x^{4} - 6 x^{2} + 4\right)$$ $$y = 64 a^{6} - x^{12}$$ ⇒ $$\left(4 a^{2} - x^{4}\right) \left(16 a^{4} + 4 a^{2} x^{4} + x^{8}\right)$$ $$y = x^{4} - 4$$ ⇒ $$\left(x^{2} - 2\right) \left(x^{2} + 2\right)$$ $$y = 4 a^{6} - 1$$ ⇒ $$\left(2 a^{3} - 1\right) \left(2 a^{3} + 1\right)$$ $$y = b^{4} - 9$$ ⇒ $$\left(b^{2} - 3\right) \left(b^{2} + 3\right)$$ $$y = 27 b^{6} + 8$$ ⇒ $$\left(3 b^{2} + 2\right) \left(9 b^{4} - 6 b^{2} + 4\right)$$ $$y = 4 x^{6} - 9 y^{12}$$ ⇒ $$\left(2 x^{3} - 3 y^{6}\right) \left(2 x^{3} + 3 y^{6}\right)$$ $$y = 27 a^{9} + 64$$ ⇒ $$\left(3 a^{3} + 4\right) \left(9 a^{6} - 12 a^{3} + 16\right)$$ $$y = - 8 a^{6} + b^{3}$$ ⇒ $$\left(- 2 a^{2} + b\right) \left(4 a^{4} + 2 a^{2} b + b^{2}\right)$$ $$y = 27 b^{6} + 64$$ ⇒ $$\left(3 b^{2} + 4\right) \left(9 b^{4} - 12 b^{2} + 16\right)$$ $$y = b^{4} - 1$$ ⇒ $$\left(b^{2} - 1\right) \left(b^{2} + 1\right)$$ $$y = c^{6} - 8$$ ⇒ $$\left(c^{2} - 2\right) \left(c^{4} + 2 c^{2} + 4\right)$$ $$y = - a^{8} + 16 c^{4}$$ ⇒ $$\left(- a^{4} + 4 c^{2}\right) \left(a^{4} + 4 c^{2}\right)$$ $$y = 27 b^{9} + x^{18}$$ ⇒ $$\left(3 b^{3} + x^{6}\right) \left(9 b^{6} - 3 b^{3} x^{6} + x^{12}\right)$$ $$y = 4 a^{4} - 9 x^{8}$$ ⇒ $$\left(2 a^{2} - 3 x^{4}\right) \left(2 a^{2} + 3 x^{4}\right)$$ $$y = 9 b^{4} - 4$$ ⇒ $$\left(3 b^{2} - 2\right) \left(3 b^{2} + 2\right)$$ $$y = 8 b^{3} - 27$$ ⇒ $$\left(2 b - 3\right) \left(4 b^{2} + 6 b + 9\right)$$ $$y = 64 c^{18} + 27 y^{9}$$ ⇒ $$\left(4 c^{6} + 3 y^{3}\right) \left(16 c^{12} - 12 c^{6} y^{3} + 9 y^{6}\right)$$ $$y = - 16 b^{12} + 9 c^{6}$$ ⇒ $$\left(- 4 b^{6} + 3 c^{3}\right) \left(4 b^{6} + 3 c^{3}\right)$$ $$y = c^{12} + 8 y^{6}$$ ⇒ $$\left(c^{4} + 2 y^{2}\right) \left(c^{8} - 2 c^{4} y^{2} + 4 y^{4}\right)$$ $$y = - 27 c^{12} + x^{6}$$ ⇒ $$\left(- 3 c^{4} + x^{2}\right) \left(9 c^{8} + 3 c^{4} x^{2} + x^{4}\right)$$
Problem Set Filters:

### Power

Reshuffle Questions