Question

if you got:
zeros: $x = \\{-4,2\\}$

now find:
use the list of linear factors to
the right to identify the two
linear factors whose product is:
$x^2 + x - 12$
a) $(x+1)$
m) $(x+15)$
b) $(x-1)$
n) $(x-15)$
c) $(x+2)$
o) $(x+10)$
d) $(x-2)$
p) $(x-10)$
e) $(x+3)$
q) $(x+12)$
f) $(x-3)$
r) $(x-12)$
g) $(x+4)$
s) $(x+20)$
h) $(x-4)$
t) $(x-20)$
i) $(x+5)$
u) $(x+30)$
j) $(x-5)$
v) $(x-30)$
k) $(x+6)$
w) $(x+60)$
l) $(x-6)$
x) $(x-60)$

Explanation:

Step1: Recall factor from zero

If \( x = a \) is a zero, then \( (x - a) \) is a factor. Given zeros \( x=-4 \) and \( x = 2 \), the factors are \( (x + 4) \) (since \( x - (-4)=x + 4 \)) and \( (x - 2) \).

Step2: Multiply the factors

Multiply \( (x + 4) \) and \( (x - 2) \):

$$
LATEXBLOCK0
$$

Wait, maybe we should factor \( x^{2}+x - 12 \) directly. Let's find two numbers that multiply to -12 and add to 1. The numbers are 4 and -3. So \( x^{2}+x - 12=(x + 4)(x - 3) \)? No, wait \( 4\times(-3)=-12 \) and \( 4+(-3)=1 \). Wait, no, earlier zeros were -4 and 2. Wait, maybe I made a mistake. Wait, if zeros are -4 and 2, then the quadratic is \( (x + 4)(x - 2)=x^{2}+2x - 8 \), but the quadratic here is \( x^{2}+x - 12 \). Wait, maybe the problem is to factor \( x^{2}+x - 12 \). Let's find two numbers that multiply to -12 and add to 1. The numbers are 4 and -3? No, 4 + (-3)=1? 4-3=1, yes. So \( x^{2}+x - 12=(x + 4)(x - 3) \)? Wait, no, \( (x + 4)(x - 3)=x^{2}+x - 12 \). Wait, but the zeros given are -4 and 2. Wait, maybe there is a mix - up. Wait, let's check the factors from the list. The list has \( (x + 4) \) (option G) and \( (x - 3) \) (option F)? Wait, no, let's re - factor \( x^{2}+x - 12 \). Let's use the zero method. Set \( x^{2}+x - 12 = 0 \). Solve for x: \( x=\frac{-1\pm\sqrt{1 + 48}}{2}=\frac{-1\pm7}{2} \). So \( x=\frac{-1 + 7}{2}=3 \) and \( x=\frac{-1-7}{2}=-4 \). Ah! So the zeros are \( x=-4 \) and \( x = 3 \). Wait, the initial zeros given were \( \{-4,2\} \), maybe that's a different problem. But for \( x^{2}+x - 12 \), the zeros are \( x=-4 \) and \( x = 3 \), so the factors are \( (x + 4) \) and \( (x - 3) \). But in the list, \( (x + 4) \) is option G and \( (x - 3) \) is option F. Wait, but maybe I made a mistake earlier. Let's check the product of \( (x + 4) \) and \( (x - 3) \): \( (x + 4)(x - 3)=x^{2}+x - 12 \), which matches. But the initial zeros were \( \{-4,2\} \), maybe that's a different quadratic. But the problem says "use the list of linear factors to identify the two linear factors whose product is \( x^{2}+x - 12 \)". Let's list the factors:
We need two factors \( (x + a) \) and \( (x + b) \) such that \( (x + a)(x + b)=x^{2}+x - 12 \), so \( a + b = 1 \) and \( ab=-12 \). Solving, we get \( a = 4 \), \( b=-3 \) (since \( 4+(-3)=1 \) and \( 4\times(-3)=-12 \)). So the factors are \( (x + 4) \) (option G) and \( (x - 3) \) (option F). Wait, but in the initial zeros, it was \( x=-4 \) and \( x = 2 \), but maybe that's a different part. Let's go back. The problem says "If you got: Zeros: \( x=\{-4,2\} \), Now find: Use the list of linear factors to identify the two linear factors whose product is \( x^{2}+x - 12 \)". Wait, maybe there is a mistake in the problem or my understanding. Wait, let's multiply \( (x + 4) \) and \( (x - 2) \): \( x^{2}+2x - 8 \), not \( x^{2}+x - 12 \). So maybe the zeros given are for a different quadratic, and we need to factor \( x^{2}+x - 12 \) regardless. So the factors of \( x^{2}+x - 12 \) are \( (x + 4) \) and \( (x - 3) \) (since \( (x + 4)(x - 3)=x^{2}+x - 12 \)). In the list, \( (x + 4) \) is option G and \( (x - 3) \) is option F. But let's check again. Wait, maybe I miscalculated the sum. Let's find two numbers that multiply to -12 and add to 1. 4 and -3: 4*(-3)=-12, 4 + (-3)=1. Yes. So the factors are \( (x + 4) \) and \( (x - 3) \).

Answer:

G. \( (x + 4) \) and F. \( (x - 3) \) (Wait, but in the list, let's check the options again. The options are: A) \( (x + 1) \), B) \( (x - 1) \), C) \( (x + 2) \), D) \( (x - 2) \), E) \( (x + 3) \), F) \( (x - 3) \), G) \( (x + 4) \), H) \( (x - 4) \), I) \( (x + 5) \), J) \( (x - 5) \), K) \( (x + 6) \), L) \( (x - 6) \), M) \( (x + 15) \), N) \( (x - 15) \), O) \( (x + 10) \), P) \( (x - 10) \), Q) \( (x + 12) \), R) \( (x - 12) \), S) \( (x + 20) \), T) \( (x - 20) \), U) \( (x + 30) \), V) \( (x - 30) \), W) \( (x + 60) \), X) \( (x - 60) \). So \( (x + 4) \) is G and \( (x - 3) \) is F. So the two factors are G. \( (x + 4) \) and F. \( (x - 3) \)