Solve the following algebraically
$\displaystyle a)x^3+5x^2=2x+6$
$\displaystyle b)x^4-13x^2=-36$
a) $\displaystyle x^3 + 5x^2 - 2x - 6 = 0$
Notice that $\displaystyle f(-1) = (-1)^3 + 5(-1)^2 - 2(-1) - 6 = -1 + 5 + 2 - 6 = 0$
So by the factor and remainder theorems, $\displaystyle x + 1$ is a factor.
Using long division, you should find that
$\displaystyle f(x) = (x + 1)(x^2 + 4x - 6) = 0$
By the null factor law
$\displaystyle x + 1 = 0$ or $\displaystyle x^2 + 4x - 6 = 0$
The first equation is easy to solve, the second is solved using the Quadratic Formula.
So $\displaystyle x = -2 - \sqrt{10}, x = -1, x = -2 + \sqrt{10}$ are the solutions.
b) $\displaystyle x^4 - 13x^2 + 36 = 0$
Notice that if I write $\displaystyle X = x^2$ the equation becomes
$\displaystyle X^2 - 13X + 36 = 0$
$\displaystyle (X - 4)(X - 9) = 0$
$\displaystyle X = 4$ or $\displaystyle X = 9$.
Since $\displaystyle X = x^2$
$\displaystyle x^2 = 4$ or $\displaystyle x^2 = 9$
So $\displaystyle x = -3, x = -2, x = 2, x = 3$ are the solutions.