#### Answer

a) 5x^2+27x+10=(5x+2)(x+5)
b) 4x^2+12x+5=(2x+1)(2x+5).

#### Work Step by Step

a) Firstly factor $5x^2$. Factors of $5$ are $1$ and $5$ so we can write
$$5x^2=5x\cdot x$$
In factorizable trinomial's we could factor
$$5x^2+27x+10 = (5x+\square)(x+\square).$$
Now let's see how can we factorize $10$:
$$10=10\cdot 1, \quad 10 = 2\cdot 5,\quad 10=5\cdot2,\quad 10=1\cdot10.$$
Now let's plug in these factors to see do we get the correct middle term.
With plugging $10$ and $1$ we get
$$(5x+10)(x+1)=5x^2+5x+10x+10 = 5x^2+15x+10$$
which does not give the correct middle term.
With plugging $5$ and $2$ we get
$$(5x+2)(x+5)=5x^2+2x+25x+10=5x^2+27x+10$$
which gives the correct middle term so the answer is
$$5x^2+27x+10=(5x+2)(x+5).$$
b) Firstly factor $4x^2$:
$$4x^2 = 2x\cdot2x,\quad 4x^2=4x\cdot x,\quad 4x^2=x\cdot 4x$$
We will try $4x^2=2x\times2x$. Then we will write
$$(2x+\square)(2x+\square).$$
Now the factors of the last term $5$ are
$$5=5\times1,\quad 5=1\times5.$$
With plugging in $5$ and $1$ we get
$$(2x+1)(2x+5)=4x^2+2x+10x+5=4x^2+12x+5$$
which gives all the correct terms so the answer is
$$4x^2+12x+5=(2x+1)(2x+5).$$