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 way
d. 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 problem
a. 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 way
d. 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)]
