134k views
0 votes
PLS HELP I NEED THIS DONE

How to factor these so that they are in parenthesis?

x^2 - 7x + 12
x^2 + 19x + 60
x^2 + 5x - 24

User Arvin
by
6.6k points

1 Answer

5 votes

Answer:

1.(x - 4) (x - 3)

2.(x + 4) (x + 15)

3.(x - 3) (x + 8)

Explanation:

1. x^2 - 7x + 12:

To solve find two factors that multiplies to 12 and adds up to -7

so two factors of 12 which add up to -7 are: -3 and -4

cause -3 x -4 is 12 and -3 + -4 is -7 we can factor using those two numbers.

then we split our -7x into two terms joining our two factors -3 and -4 with a variable which gives -3x - 4x. once we have done that we can then factorize by grouping.

x^2 - 3x - 4x + 12

(x^2 - 3x) - (4x + 12)

we then find the h.c.f of each group of terms and we can factor.

x(x - 3) - 4(x - 3) = (x - 4) (x - 3)

now we can follow the same procedure to solve the other problems or factorize.

2.x^2 + 19x + 60

let's try to find pairs of factors which multiply to 60.

60 = (5, 12), (6, 10), (15, 4)

once we find a pair of factors which adds to 19 which the pair is (15, 4)

we split the 19x again so we can get a four term expression:

x^2 + 15x + 4x + 60

let's group

(x^2 + 15x) + (4x + 60)

hcf and factorize

x(x + 15) + 4(x + 15)

now factorize completely

(x + 4)(x + 15)

3.x^2 + 5x - 24

pairs of factors of -24: (6, -4), (8, -3), (12, -2)

pair which adds to 5: (8, -3)

x^2 + 8x - 3x - 24

(x^2 + 8x) - (3x - 24)

HCF and Factorize completely.

x(x + 8) -3(x + 8) = (x - 3)(x + 8)

User Gerard Sexton
by
8.4k points