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

Chapter 6 - Section 6.3 - Exercise Set: 89 (Answer) The perimeter of the triangle is $(4x^2 + 21x + 5)$ and $(4x^2 + 21x + 5)$ = $(x + 5)(4x + 1)$
Chapter 6 - Section 6.3 - Exercise Set: 89 (Solution) The perimeter of the triangle is $(3x^2 + 1) + (6x + 4) + (x^2 + 15x)$ $= (4x^2 + 21x + 5)$ (Combining like terms) To factorize : $(4x^2 + 21x + 5)$ Factors of $4x^2$ $\Rightarrow$ $x\cdot 4x$ or $4x\cdot x$ or $2x\cdot 2x$ Factors of 5 $\Rightarrow$ $1\cdot 5$ or $5\cdot 1$ Try combination of those factors $(x + 1)(4x + 5) = 4x^2 + 9x + 5$ (Incorrect middle term) $(x + 5)(4x + 1) = 4x^2 + 21x + 5$ (Correct middle term) $(2x + 5)(2x + 1) = 4x^2 + 12x + 5$ (Incorrect middle term) Thus, $(4x^2 + 21x + 5)$ = $(x + 5)(4x + 1)$