How many factors in the expression 8(x+4)(y+4)(z²+4z+7) have exactly 2 terms?

Answers

The answer is 2 factors.

There are four factors:
8                       - 1 term (8)
(x + 4)               - 2 terms (x and 4)
(y + 4)               - 2 terms (y and 4)
(z² + 4z + 7)      - 3 terms (z, 4z, and 7)
It helps to first identify the factors. They are:

- 8
- (x + 4)
- (y + 4)
- (z² + 4z + 7)

Terms are separated by "+" and "-" signs, so that eliminates the first factor, which only has one term. The fourth factor has two "+" signs, and as you can see, it has three terms; that one is out as well. The last two factors both have only one "+" sign, and they both have two terms. Therefore...

Two factors in the expression 8(x + 4)(y + 4)(z² + 4z + 7) have exactly two terms.

There are 2 Factors which has 2 terms.

Step-by-step explanation:

Given: Expression = 8(x+4)(y+4)(z^2+4z+7)

To find: No. of factors with 2 terms

It's clear, Given expression can not be further factorized.

So, we have 4 factors in it.

Factor 1 :

8

it has 1 term.i.e., 8 itself.

Factor 2:

x + 4

It has 2 terms .i.e., x & 4

Factor 3:

y + 4

It has 2 terms.i.e., y & 4

Factor 4:

z^2+4z+7

It has 3 terms.i.e., z^2, 4x & 7

Therefore, There are 2 Factors which has 2 terms.

2 factors.
i think so
(x+4) and (y+4) both have two terms..terms are separated by addition/subtraction operators
2 the expression has 2 binomials


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