## Algebra: A Combined Approach (4th Edition)

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 $$4x^2+12x+5=(2x+1)(2x+5).$$