Match the polynomials with their factors. 2a2 + 5a − 3 (a + 1)(2a − 3) 2a2 − a − 3 (a − 1)(2a + 3) 2a2 − 5a − 3 (2a − 1)(a + 3) 2a2 + a − 3 (2a + 1)(a − 3)

2a² + 5a - 3 = (2a - 1)(a + 3)

2a² - a - 3 = (2a - 3)(a + 1)

2a² - 5a - 3 = (2a + 1)(a - 3)

2a² + a - 3 = (2a + 3)(a - 1)

Step-by-step 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

∵ 2a² + 5a - 3

∴ c = -3 ⇒ -ve term

∴ r , s have different sign

∵ 3 = 1 × 3 then r = 1 , s = 3

∵ a = 2

∵ a = h × k

∵ 2 = 2 × 1 then h = 2 , k = 1

∵ rk = 1

∵ sh = 6

∴ sh - rk = 5 ⇒ same value of b

∵ (hx - r)(kx + s)

∴ 2a² + 5a - 3 = (2a - 1)(a + 3)

∵ 2a² - a - 3

∴ c = -3 ⇒ -ve term

∴ r , s have different sign

∵ 3 = 3 × 1 then r = 3 , s = 1

∵ a = 2

∵ a = h × k

∵ 2 = 2 × 1 then h = 2 , k = 1

∵ rk = 3

∵ sh = 2

∴ sh - rk = -1 ⇒ same value of b

∵ (hx - r)(kx + s)

∴ 2a² - a - 3 = (2a - 3)(a + 1)

∵ 2a² - 5a - 3

∴ c = -3 ⇒ -ve term

∴ r , s have different sign

∵ 3 = 1 × 3 then r = 1 , s = 3

∵ a = 2

∵ a = h × k

∵ 2 = 2 × 1 then h = 2 , k = 1

∵ rk = 1

∵ sh = 6

∴ rk - hs = -5 ⇒ same value of b

∵ (hx + r)(kx - s)

∴ 2a² - 5a - 3 = (2a + 1)(a - 3)

∵ 2a² + a - 3

∴ c = -3 ⇒ -ve term

∴ r , s have different sign

∵ 3 = 3 × 1 then r = 3 , s = 1

∵ a = 2

∵ a = h × k

∵ 2 = 2 × 1 then h = 2 , k = 1

∵ rk = 3

∵ sh = 2

∴ rk - sh = 1 ⇒ same value of b

∵ (hx + r)(kx - s)

∴ 2a² + a - 3 = (2a + 3)(a - 1)

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

the answers c