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Doss [256]
3 years ago
6

How to factorise by grouping for 7a, d and h

Mathematics
2 answers:
asambeis [7]3 years ago
7 0

Step-by-step explanation:

7a

x2 + 4x + ax + 4a

x2 + ax + 4x + 4a

x (x + a) + 4 (x + a)

:(x + 4)(x+a)

7d

x2+2x-ax-2a

x2-ax+2x-2a

x (x-a)+2 (x-a)

:(x+2)(x-a)

7h

x2-2bx-5x+10b

x2-5x-2bx+10b

x (x-5)-2b(x-5)

:(x-2b)(x-5)

Maksim231197 [3]3 years ago
5 0

Answer:

You have to take out the common terms :

Question A,

{x}^{2}  + 4x + ax + 4a

= x(x + 4) + a(x + 4)

= (x + a)(x + 4)

Question D,

{x}^{2}  + 2x - ax - 2a

= x(x  +  2) - a(x + 2)

= (x - a)(x + 2)

Question H,

{x}^{2}  - 2bx - 5x + 10b

= x(x - 2b) - 5(x - 2b)

= (x - 5)(x - 2b)

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Download

GET

Algebra

Factor

Factor the polynomial.

(

x

+

3

)

(

x

+

5

)

Tap to view FREE steps...

()|[]√≥

789≤

456/^×>∩∪

123-+÷<π∞

,0.%=

Factor

x

2

+

8

x

+

15

+

0

Regroup terms.

x

2

+

0

+

8

x

+

15

Add

x

2

and

0

.

x

2

+

8

x

+

15

Factor

x

2

+

8

x

+

15

using the AC method.

Tap for fewer steps...

Consider the form

x

2

+

b

x

+

c

. Find a pair of integers whose product is

c

and whose sum is

b

. In this case, whose product is

15

and whose sum is

8

.

3

,

5

Write the factored form using these integers.

(

x

+

3

)

(

x

+

5

)

4 0
3 years ago
Read 2 more answers
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