MATH SOLVE

3 months ago

Q:
# 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)

Accepted Solution

A:

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)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)