MATH SOLVE

3 months ago

Q:
# Help solve these please show little examples of possible

Accepted Solution

A:

Answer:a. x² + x - 30 = (x + 6)(x - 5)b. -3x² + 23x - 14 = -[(3x - 2)(x - 7)]c. 2x² - 5x + 4 can not factorize by this wayd. 6x² + 10x - 24 = 2[(3x - 4)(x + 3)]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 problema. x² + x - 30∵ ax² + bx + c∴ a = 1 , b = 1 , c = -30∵ c is negative∴ r and s have different signs∵ a = h × k∵ 1 = 1 × 1∴ h = 1 , k = 1∵ c = r × s∵ c = -30∴ r × s = -30∵ 6 × -5 = -30∴ r = 6 , s = -5∴ hs = 6∴ rk = -5∵ hs - rk = 6 - 5 = 1 ⇒ same value of b∴ (x + 6)(x - 5)* x² + x - 30 = (x + 6)(x - 5)b. -3x² + 23x - 14 ⇒ take -1 as a common factor∴ -(3x² - 23x + 14)∵ ax² + bx + c∴ a = 3 , b = -23 , c = 14∵ c is positive ∴ r and s have same sign (-ve) because b is negative∵ a = h × k∵ 3 = 3 × 1∴ h = 3 , k = 1∵ c = r × s∵ 14 = 2 × 7∴ r = 2 , s = 7∴ hs = 3 × 7 = 21∴ rk = 2 × 1 = 2∵ hs + rk = 21 + 2 = 23 ⇒ same value of b∴ (3x - 2)(x - 7)* -3x² + 23x - 14 = -[(3x - 2)(x - 7)]c. 2x² - 5x + 4∵ ax² + bx + c∴ a = 2 , b = -5 , c = 4∵ c is positive ∴ r and s have same sign (-ve) because b is negative∵ a = h × k∵ 2 = 2 × 1∴ h = 2 , k = 1∵ c = r × s∵ 4 = 2 × 2∴ r = 2 , s = 2∴ hs = 2 × 2 = 4∴ rk = 2 × 1 = 2∵ hs + rk = 4 + 2 = 6 ⇒ not same value of b∴ We can not factorize it* 2x² - 5x + 4 can not factorize by this wayd. 6x² + 10x - 24 ⇒ take 2 as a common factor∴ 2(3x² + 5x - 12)∵ ax² + bx + c∴ a = 3 , b = 5 , c = -12∵ c is negative∴ r and s have different signs∵ a = h × k∵ 3 = 3 × 1∴ h = 3 , k = 1∵ c = r × s∵ -12 = -4 × 3∴ r = -4 , s = 3∴ hs = 3 × 3 = 9∴ rk = -4 × 1 = -4∵ hs - rk = 9 - 4 = 5 ⇒ same value of b∴ (3x - 4)(x + 3)* 6x² + 10x - 24 = 2[(3x - 4)(x + 3)]