Algebra — practice questions

We are asked to solve $|3x - 1| = |x + 4|$. A robust method is to square both sides of the equation.

$(3x - 1)^2 = (x + 4)^2$

Expand both sides: $(3x - 1)(3x - 1) = (x + 4)(x + 4)$ $9x^2 - 6x + 1 = x^2 + 8x + 16$

Now, rearrange to form a quadratic equation equal to zero: $9x^2 - x^2 - 6x - 8x + 1 - 16 = 0$ $8x^2 - 14x - 15 = 0$

We can solve this quadratic equation. We look for two numbers that multiply to $8 \times (-15) = -120$ and add to $-14$. These numbers are $-20$ and $6$. $8x^2 - 20x + 6x - 15 = 0$ $4x(2x - 5) + 3(2x - 5) = 0$ $(4x + 3)(2x - 5) = 0$

This gives two possible solutions: $4x + 3 = 0 \implies x = -\frac{3}{4}$ $2x - 5 = 0 \implies x = \frac{5}{2}$

The solutions are $x = -\frac{3}{4}$ and $x = \frac{5}{2}$.

(i) To show $(x-2)$ is a factor, we use the Factor Theorem and evaluate $p(2)$: $p(2) = 2(2)^3 + 3(2)^2 - 18(2) + 8$ $= 2(8) + 3(4) - 36 + 8$ $= 16 + 12 - 36 + 8$ $= 28 - 36 + 8 = -8 + 8 = 0$ Since $p(2)=0$, $(x-2)$ is a factor of $p(x)$.

(ii) Since $(x-2)$ is a factor, we can write $p(x)$ in the form $(x-2)(Ax^2 + Bx + C)$. By comparing the coefficient of $x^3$, we see $A=2$. By comparing the constant term, we see $-2C = 8$, so $C=-4$. So, $p(x) = (x-2)(2x^2 + Bx - 4)$. To find $B$, we can compare the $x^2$ coefficient. The $x^2$ term is formed by $x(Bx) + (-2)(2x^2) = (B-4)x^2$. We know the coefficient of $x^2$ in $p(x)$ is 3. So, $B-4 = 3 \implies B=7$.

Thus, $p(x) = (x-2)(2x^2 + 7x - 4)$. Now we factorise the quadratic $2x^2 + 7x - 4$. We look for two numbers that multiply to $2 \times (-4) = -8$ and add to $7$. These are $8$ and $-1$. $2x^2 + 8x - x - 4 = 2x(x+4) - 1(x+4) = (2x-1)(x+4)$.

Therefore, the complete factorisation is $p(x) = (x-2)(x+4)(2x-1)$.