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

Practice 4a (Answer) $q^{2}$ - 3$q$ - 40 = ($q$ - 8)($q$ + 5) Practice 4b (Answer) $y^{2}$ + 2$y$ - 48 = ($y$ - 6)($y$ + 8)
Practice 4a (Solution) Factor : $q^{2}$ - 3$q$ - 40 To begin by writing the first terms of the binomial factors ($q$ + $\triangle$)($q$ + $\square$) Next, to look for two numbers whose product is -40 and whose sum is -3. As the two numbers must have a negative product, pairs of factors with opposite signs of -40 are to be investigated only. Factors of -40 $\Longleftrightarrow$ Sum of Factors -1,40 $\Longleftrightarrow$ 39 -40,1 $\Longleftrightarrow$ -39 -2,20 $\Longleftrightarrow$ 18 -20,2 $\Longleftrightarrow$ -18 -4,10 $\Longleftrightarrow$ 6 -10,4 $\Longleftrightarrow$ -6 -5,8 $\Longleftrightarrow$ 3 -8,5 $\Longleftrightarrow$ -3 (Correct sum, so the numbers are -8 and 5) Thus, $q^{2}$ - 3$q$ - 40 = ($q$ - 8)($q$ + 5) Practice 4b (Solution) Factor : $y^{2}$ + 2$y$ - 48 To begin by writing the first terms of the binomial factors ($y$ + $\triangle$)($y$ + $\square$) Next, to look for two numbers whose product is -48 and whose sum is +2. As the two numbers must have a negative product, pairs of factors with opposite signs of -48 are to be investigated only. Factors of -48 $\Longleftrightarrow$ Sum of Factors -1,48 $\Longleftrightarrow$ 47 -48,1 $\Longleftrightarrow$ -47 -2,24 $\Longleftrightarrow$ 22 -24,2 $\Longleftrightarrow$ -22 -3,16 $\Longleftrightarrow$ 13 -16,3 $\Longleftrightarrow$ -13 -4,12 $\Longleftrightarrow$ 8 -12,4 $\Longleftrightarrow$ -8 -6,8 $\Longleftrightarrow$ 2 (Correct sum, so the numbers are -6 and 8) -8,6 $\Longleftrightarrow$ -2 Thus, $y^{2}$ + 2$y$ - 48 = ($y$ - 6)($y$ + 8)