Factor 2x2 + 7x + 3. (2x + 2)(x + 1) (2x + 3)(x + 1) (2x + 1)(x + 3)

(2x + 1)(x + 3)

It is probably easier to try the answer choices than to try to factor the expression yourself.

(2x + 2)(x + 1) = 2x² +4x +2

(2x + 3)(x + 1) = 2x² +5x +3

(2x + 1)(x + 3) = 2x² +7x +3 . . . . . correct choice

Constructed solution

If you want to factor this yourself, you can look for factors of "ac" that add to give "b". That is, you want factors of 2·3 = 6 that add up to give 7. You don't have to look very far.

... 6 = 1·6 = 2·3 . . . . . . the first factor pair adds to give 7

Now, rewrite the x term using the sum of these numbers.

... 2x² +(1 +6)x +3

... 2x +x +6x +3 . . . . eliminate parentheses

... (2x +x) +(6x +3) . . . . group pairs of terms

... x(2x +1) +3(2x +1) . . . . factor each pair

... (x +3)(2x +1) . . . . . . matches the last selection

ion

step-by-step explanation:

ion ion ion ion ion ion ion ion ion ion ion

< 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)