What is the factored form of the trinormial x^2+11x+28=0

x={-7,-4}

Step-by-step explanation: Simplifying

x2 + 11x + 28 = 0

Reorder the terms:

28 + 11x + x2 = 0

Solving

28 + 11x + x2 = 0

Solving for variable 'x'.

Factor a trinomial.

(7 + x)(4 + x) = 0

Subproblem 1

Set the factor '(7 + x)' equal to zero and attempt to solve:

Simplifying

7 + x = 0

Solving

7 + x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-7' to each side of the equation.

7 + -7 + x = 0 + -7

Combine like terms: 7 + -7 = 0

0 + x = 0 + -7

x = 0 + -7

Combine like terms: 0 + -7 = -7

x = -7

Simplifying

x = -7

Subproblem 2

Set the factor '(4 + x)' equal to zero and attempt to solve:

Simplifying

4 + x = 0

Solving

4 + x = 0

Move all terms containing x to the left, all other terms to the right.

Add '-4' to each side of the equation.

4 + -4 + x = 0 + -4

Combine like terms: 4 + -4 = 0

0 + x = 0 + -4

x = 0 + -4

Combine like terms: 0 + -4 = -4

x = -4

Simplifying

x = -4

Solution

x = {-7, -4}

a

step-by-step explanation:

rurbrbrjrnrbjr

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)