Math Problems

MathematicsPrecalculusCubic Equations → Finding Factors of Basic Polynomials

Factor the polynomial

## Questions

 $$y = 27 a^{6} + 1$$ $$y = 4 a^{6} - b^{12}$$ $$y = c^{9} - 8$$ $$y = x^{6} - 64$$ $$y = 8 b^{6} - 27$$ $$y = 8 b^{6} - 1$$ $$y = 16 a^{2} - 9 x^{4}$$ $$y = a^{3} - 8 y^{6}$$ $$y = 8 a^{3} - 1$$ $$y = 4 c^{4} - 9$$ $$y = - b^{18} + 8 c^{9}$$ $$y = b^{6} - 64$$ $$y = 27 x^{9} - 8$$ $$y = 8 x^{6} - 27 y^{12}$$ $$y = 27 b^{6} + 64$$ $$y = b^{6} + 64$$ $$y = c^{3} + 8$$ $$y = c^{2} - 16$$ $$y = 64 c^{9} + 27 x^{18}$$ $$y = 27 x^{6} + 8 y^{12}$$ $$y = 27 x^{9} - 1$$ $$y = a^{9} + 64 c^{18}$$ $$y = 27 a^{6} - 64$$ $$y = 64 c^{12} + 27 y^{6}$$ $$y = 8 c^{6} - 1$$ $$y = 8 x^{6} + 27$$ $$y = c^{6} - 8$$ $$y = a^{9} - 27$$

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

### Power

Reshuffle Questions