We want to find the least common multiple of the given pair of polynomials.

9 x^{2} - 16 and 3 x^{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.

9 x^{2} - 16

WritePow

Write as a power

3^{2} x^{2} - 4^{2}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

(3 x)^{2} - 4^{2}

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

(3 x + 4) (3 x - 4)

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.

3 x^{2} + x - 4 \Leftrightarrow 3 x^{2} + 1 x + (- 4)

Let's use the Quadratic Formula to factor our polynomial!

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 1 \pm 1 ^{2} - 4 ( 3 ) ( - 4 )}{2 ( 3 )}

▼

Simplify right-hand side

CalcPow

Calculate power

x = \frac{- 1 \pm 1 - 4 ( 3 ) ( - 4 )}{2 ( 3 )}

Multiply

x = \frac{- 1 \pm 1 + 4 8}{6}

AddTerms

Add terms

x = \frac{- 1 \pm 4 9}{6}

CalcRoot

Calculate root

x = \frac{- 1 \pm 7}{6}

Knowing that

x = \frac{- 1 + 7}{6} = 1

and

x = \frac{- 1 - 7}{6} = - \frac{4}{3}

are the zeros of the trinomial, we can write the factored form of the quadratic expression.

3 (x - 1) (x - (- \frac{4}{3})) ⇕ (x - 1) (3 x + 4)

We can now write both expressions as the product of their prime factors.

9 x^{2} - 16 3 x^{2} + x - 4 = (3 x + 4) (3 x - 4) = (x - 1) (3 x + 4)

Finally, we write the product of the prime factors, each raised to the greatest power that occurs in their expression.

(3 x + 4) (3 x - 4) (x - 1)

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