Solve the equation x4 + 2x3 − 7x2 − 8x + 12 = 0

Let $P(x) = x^4 + 2x^3 - 7x^2 - 8x + 12$.
We test integer divisors of the constant term 12 ($\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$).
For $x=1$:
$P(1) = (1)^4 + 2(1)^3 - 7(1)^2 - 8(1) + 12 = 1 + 2 - 7 - 8 + 12 = 0$.
So, $x=1$ is a root, and $(x-1)$ is a factor.

Dividing $P(x)$ by $(x-1)$ gives $x^3 + 3x^2 - 4x - 12$.
So, the equation becomes $(x-1)(x^3 + 3x^2 - 4x - 12) = 0$.

Again, test integer divisors of the constant term 12.
For $x=2$:
$Q(2) = (2)^3 + 3(2)^2 - 4(2) - 12 = 8 + 3(4) - 8 - 12 = 8 + 12 - 8 - 12 = 0$.
So, $x=2$ is a root, and $(x-2)$ is a factor of $Q(x)$.

Dividing $Q(x)$ by $(x-2)$ gives $x^2 + 5x + 6$.
So, the equation becomes $(x-1)(x-2)(x^2 + 5x + 6) = 0$.

This quadratic factors as $(x+2)(x+3)$.
So, the equation is $(x-1)(x-2)(x+2)(x+3) = 0$.

$x-1 = 0 \implies x = 1$
$x-2 = 0 \implies x = 2$
$x+2 = 0 \implies x = -2$
$x+3 = 0 \implies x = -3$

The solutions are:

: Geometry Problem

Given:

• Triangle ABC is a right-angled triangle at B.
• Length of side AB = 5 cm.
• Length of side BC = 12 cm.
• Point D is on AC such that BD is perpendicular to AC.

Required: Find the length of BD.

Since $\triangle ABC$ is a right-angled triangle at B, we have:

The area of a triangle can be calculated using the formula $\frac{1}{2} \times \text{base} \times \text{height}$.
We can calculate the area of $\triangle ABC$ in two ways:

1. Using AB as base and BC as height (since $\angle B = 90^\circ$): $$\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times BC$$ $$\text{Area}(\triangle ABC) = \frac{1}{2} \times 5 \text{ cm} \times 12 \text{ cm}$$ $$\text{Area}(\triangle ABC) = \frac{1}{2} \times 60 \text{ cm}^2$$ $$\text{Area}(\triangle ABC) = 30 \text{ cm}^2$$
2. Using AC as base and BD as height (since BD is perpendicular to AC): $$\text{Area}(\triangle ABC) = \frac{1}{2} \times AC \times BD$$

Substitute the value of AC:

The length of BD is: