Exercise 14 Page 388

We want to find the least common multiple of the given pair of polynomials. 9x^2-16 and 3x^2+x-4 To do so, we first need to find the prime factors of each expression. Let's start by factoring the first one. Note that this polynomial can be rewritten as a difference of squares. We will use this fact to factor it. Let's now factor the second expression. The degree of the second polynomial is 2 and the polynomial does not create a perfect square trinomial. Therefore, we can use the Quadratic Formula to help us factor. We will start by identifying the values of a, b and c in our polynomial. 3x^2+x-4 ⇔ 3x^2+ 1x+( -4) Let's use the Quadratic Formula to factor our polynomial!

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 3)( - 4))/2( 3)

▼

Simplify right-hand side

CalcPow

Calculate power

x=- 1±sqrt(1-4(3)(- 4))/2(3)

Multiply

Multiply

x=- 1±sqrt(1+48)/6

AddTerms

Add terms

x=- 1± sqrt(49)/6

CalcRoot

Calculate root

x=- 1± 7/6

Knowing that x= - 1+76= 1 and x= - 1-76=- 43 are the zeros of the trinomial, we can write the factored form of the quadratic expression. 3(x- 1)(x-(-4/3)) ⇕ (x-1)(3x+4) We can now write both expressions as the product of their prime factors. 9x^2-16 &= (3x+4) (3x-4) 3x^2+x-4 &= (x-1) (3x+4) Finally, we write the product of the prime factors, each raised to the greatest power that occurs in their expression. (3x+4) (3x-4) (x-1)