1. **State the problem:** We want to find the value of $a$ such that \(x^2 - 14x + 49 = (x - a)^2\). 2. **Recall the expansion:** The right side is a perfect square trinomial, expanded as $$ (x - a)^2 = x^2 - 2ax + a^2 $$ 3. **Compare coefficients:** The left side is \(x^2 - 14x + 49\) and the right side is \(x^2 - 2ax + a^2\). Matching coefficients term-by-term: - The coefficient of $x$ is \(-14 = -2a\) - The constant term is \(49 = a^2\) 4. **Solve for $a$ from the $x$ coefficient:** $$ -14 = -2a \Rightarrow 2a = 14 \Rightarrow a = 7 $$ 5. **Check the constant term:** $$ a^2 = 7^2 = 49 $$ This matches perfectly. 6. **Conclusion:** The value of $a$ is \(\boxed{7}\). --- Next, **fully factorise the expression:** \(20x - 44 + x^2\). 1. **Rewrite the expression:** $$ x^2 + 20x - 44 $$ 2. **Look for factors of the quadratic:** We want to factor as $$ (x + m)(x + n) $$ such that $$ mn = -44 $$ and $$ m + n = 20 $$ 3. **Check possible factor pairs of 44:** Those are 1 & 44, 2 & 22, 4 & 11. None of these pairs sum to 20. 4. **Use the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{400 + 176}}{2} = \frac{-20 \pm \sqrt{576}}{2} = \frac{-20 \pm 24}{2} $$ So the roots are $$ x = \frac{-20 + 24}{2} = 2, \quad x = \frac{-20 - 24}{2} = -22 $$ 5. **Write the factorisation:** $$ x^2 + 20x - 44 = (x - 2)(x + 22) $$ ---