The factors of m2 + 12m + 35 are ( ) and ( ). nextreset

The answer is (m+5) (m+7).

[tex](m+5)(m+7)[/tex]

Step-by-step explanation:

Factor the trinomial. The factors of [tex]m^2 + 12m + 35[/tex]

In the given trinomial , the product is 35 and sum is 12

LEts find out two factors whose product is 35 and sum is 12

7 times 5 is 35

7 plus 5 is 12

Break the middle term using factors 7 and 5

[tex]m^2 + 12m + 35[/tex]

[tex]m^2 + 7m+5m + 35[/tex]

Group first two terms and last two terms

[tex](m^2 + 7m)+(5m + 35)[/tex]

Take out GCF from each group

[tex]m(m+7)+5(m + 7)[/tex]

FActor out m+7

[tex](m+5)(m+7)[/tex]

Factors are (x+7) and (x+5).

Step-by-step explanation:

The given trinomial is (m² + 12m + 35)

For factorization of any polynomial we should apply zero factor property.

if a given polynomial is in the form of (ax² + bx + c) then the zero factors will be in the form of c/a.

Here c represents the multiples of constant term and a represents the multiples of the coefficient of highest polynomial.

Now our equation is m² + 12m + 35

here c is 35 and a is 1

Now multiples of 35 are ±7, ±5 and for 1 are ±1.

Now zero factors will be c/a = ±7/±1, ±5/±1 Or 7, 5 are the zero factors of the equation

Now we can say that (x+7) and (x+5) are the zero factors of the trinomial

So the trinomial can be written as (x+7)(x+5).

m^2 + 12m + 35

We can factor using the sum-product, when the sum of 2 numbers is 12 and the product of those 2 numbers is 35.

7+5 = 12

7 x 5 = 35

So we have the numbers 7 and 5.

m^2 + 12m + 35 = (m+7)(m+5)

The answer is (m+5)(m+7)
Do you need work?

(m+7) and (m+5)

Step-by-step explanation:

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