Answer
Practice 4a (Answer)
$q^{2}$ - 3$q$ - 40
= ($q$ - 8)($q$ + 5)
Practice 4b (Answer)
$y^{2}$ + 2$y$ - 48
= ($y$ - 6)($y$ + 8)
Work Step by Step
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)
