Factorise $3x^2-14x-5$. Answer ... Hence, factorise completely $3(y+1)^2-14y-19$. Answer ...

1. Find a suitable split for the middle term

For $3x^2-14x-5$, calculate

$3\times(-5)=-15.$

We need two numbers with product $-15$ and sum $-14$:

$-15 \text{ and } 1.$

2. Split the middle term and factorise by grouping

$3x^2-14x-5=3x^2-15x+x-5$

$=3x(x-5)+1(x-5)$

$=(3x+1)(x-5)$

1. Rewrite the expression so it matches the quadratic from part (a)

Since $-14y-19=-14(y+1)-5$,
$3(y+1)^2-14y-19=3(y+1)^2-14(y+1)-5.$

2. Apply the factorisation from part (a) and simplify the factors

From part (a),
$3x^2-14x-5=(3x+1)(x-5).$
Let $x=y+1$.
$3(y+1)^2-14(y+1)-5=[3(y+1)+1][(y+1)-5]$
$=(3y+4)(y-4).$