Factorise fully:
$6(p-8)+4(p-8)^2$
The expression to factorise is \(6(p-8) + 4(p-8)^2\). We observe that both terms contain the binomial factor \((p-8)\). Additionally, both numerical coefficients, 6 and 4, share a common factor of 2. Therefore, the expression can be factored by extracting the greatest common factor of the entire expression, which is \(2(p-8)\).
We factor out \(2(p-8)\) from both parts of the expression:
\[6(p-8) + 4(p-8)^2 = 2(p-8) \cdot 3 + 2(p-8) \cdot 2(p-8)\]
By factoring out \(2(p-8)\), we get:
\[2(p-8) [3 + 2(p-8)]\]
Now, we simplify the expression inside the brackets:
\[3 + 2(p-8) = 3 + 2p - 16 = 2p - 13\]
Substituting this back, the expression becomes:
\[2(p-8)(2p-13)\]
Thus, the fully factorised form is \(2(p-8)(2p-13)\).
Factoring expressions is a foundational skill used in engineering and physics to simplify equations describing physical systems, such as finding the time at which a projectile returns to a specific height or determining the roots of a quadratic model. For example, if the height of an object is given by \(h(t) = 6(t-8) + 4(t-8)^2\), setting this to zero to find when the object is at ground level allows us to solve \(2(t-8)(2t-13) = 0\), yielding simple solutions \(t=8\) and \(t=6.5\). Mastering this allows you to efficiently solve complex algebraic models in real-world scenarios.