Answer:
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)
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)