Question

Factor completely 3y^2+7y+4

Answer 1

The factors of 3y^2 + 7y + 4 are

(y +1) and (3y + 4)

Explanation:

It is given a quadratic equation in variable y

3y^2 + 7y + 4

To find the factors using splitting method

3y^2 + 7y + 4 can be written as,

3y^2 + 7y + 4 = 3y^2 + 3y + 4y + 4

= 3y(y + 1) + 4(y + 1)

= (y + 1)(3y + 4)

Therefore the factors of 3y^2 + 7y + 4 are

(y +1) and (3y + 4)

Answer 2

3y² + 7y + 4 = (3x + 4)(x + 1)

Explanation:

* To factor a trinomial in the form ax² ± bx ± c:

- Look at the c term

# If the c term is positive

∵ c = r × s ⇒ r and s are the factors of c

∴ r and s will have the same sign (sign of b)

∵ a = h × k ⇒ h , k are the factors of a

∴ rk + hs = b

∴ (hx + r)(kx + s) ⇒ if b +ve OR (hx - r)(kx - s) ⇒ if b -ve

# If the c term is negative

∵ c = r × s ⇒ r and s are the factors of c

∴ r and s will not have the same sign

∵ a = h × k ⇒ h and k are the factors of a

∴ rk - hs = b OR hs - rk = b

(hx + r)(kx - s) OR (hx - r)(kx + s)

* Now lets solve the problem

∵ 3y² + 7y + 4

∵ ax² + bx + c

∴ a = 3 , b = 7 , c = 4

∵ a = h × k

∵ 3 = 3 × 1

∴ h = 3 , k = 1

∵ c = r × s

∵ 4 = 4 × 1

∴ r = 4 , s = 1

∵ c is positive

∴ hs + rk = b

∴ 3(1) + 4(1) = 7 ⇒ same value of b

∴ 3y² + 7y + 4 = (3x + 4)(x + 1)