Solving by factorising questions with answers

Questions

1. Question 1x² + bx + c = 0

   Solve $x^{2} + 6x - 27 = 0$.

   Show the worked solution

   1. Step 1: Factorise$(x - 3)(x + 9) = 0$
   2. Step 2: Set each factor to zero$x - 3 = 0$ or $x + 9 = 0$$x = 3$ or $x = -9$
2. Question 2ax² + bx + c = 0

   Solve $3x^{2} + 5x + 2 = 0$.

   Show the worked solution

   1. Step 1: Factorise$(3x + 2)(x + 1) = 0$
   2. Step 2: Set each factor to zero$3x + 2 = 0$ or $x + 1 = 0$$x = -\dfrac{2}{3}$ or $x = -1$
3. Question 3Rearrange first

   Solve $5x^{2} -8x = -3$.

   Show the worked solution

   1. Step 1: Make one side zero$5x^{2} - 8x + 3 = 0$
   2. Step 2: Factorise$(5x - 3)(x - 1) = 0$
   3. Step 3: Set each factor to zero$5x - 3 = 0$ or $x - 1 = 0$$x = \dfrac{3}{5}$ or $x = 1$
4. Question 4Common factor x (don't divide by x)

   Solve $3x^{2} = -2x$.

   Show the worked solution

   1. Step 1: Make one side zero$3x^{2} + 2x = 0$Dividing both sides by $x$ would lose a solution, so don't.
   2. Step 2: Factorise$x(3x + 2) = 0$
   3. Step 3: Set each factor to zero$x = 0$ or $3x + 2 = 0$$x = 0$ or $x = -\dfrac{2}{3}$
5. Question 5Expand the brackets first

   Solve $x(2x + 13) = -15$.

   Show the worked solution

   1. Step 1: Expand the bracket$2x^{2} + 13x = -15$
   2. Step 2: Make one side zero$2x^{2} + 13x + 15 = 0$
   3. Step 3: Factorise$(2x + 3)(x + 5) = 0$
   4. Step 4: Set each factor to zero$2x + 3 = 0$ or $x + 5 = 0$$x = -\dfrac{3}{2}$ or $x = -5$
6. Question 6Fractional: x + k/x

   Solve $x - \dfrac{6}{x} = -1$.

   Show the worked solution

   1. Step 1: Multiply every term by the denominator$x^{2} - 6 = -x$
   2. Step 2: Make one side zero$x^{2} + x - 6 = 0$
   3. Step 3: Factorise$(x + 3)(x - 2) = 0$
   4. Step 4: Set each factor to zero$x + 3 = 0$ or $x - 2 = 0$$x = -3$ or $x = 2$Neither solution makes a denominator zero, so both are valid.
7. Question 7Fractional: two fractions

   Solve $\dfrac{1}{x - 3} + \dfrac{6}{x + 3} = 1$.

   Show the worked solution

   1. Step 1: Multiply every term by the LCM of the denominators$1(x + 3) + 6(x - 3) = (x - 3)(x + 3)$
   2. Step 2: Expand the brackets$7x - 15 = x^{2} - 9$
   3. Step 3: Make one side zero$x^{2} - 7x + 6 = 0$
   4. Step 4: Factorise$(x - 1)(x - 6) = 0$
   5. Step 5: Set each factor to zero$x - 1 = 0$ or $x - 6 = 0$$x = 1$ or $x = 6$Neither solution makes a denominator zero, so both are valid.
8. Question 8One solution given: find k, then the other

   Given that $x = 2$ is a solution of the equation $2x^{2} + kx - 10 = 0$, find the other solution.

   Show the worked solution

   1. Step 1: Substitute x = 2 into the equation$22^{2} + k2 - 10 = 0$
   2. Step 2: Solve for k$k = 1$
   3. Step 3: Put k back and factorise: (x − 2) is one factor$2x^{2} + x - 10 = 0$$(x - 2)(2x + 5) = 0$
   4. Step 4: The other factor gives the other solution$x = -\dfrac{5}{2}$