Solve these show little examples of possible

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)]

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step-by-step explanation:

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step-by-step explanation:

because that's the answer