\( 3 x^{2}-7 x-6 \quad \) 2) \( 9 x^{4}-13 x^{2}+4 \) \( x(3 x \)

It seems you have provided two polynomial expressions: 1. \( 3x^2 - 7x - 6 \) 2. \( 9x^4 - 13x^2 + 4 \) Additionally, there is a fragment \( x(3x) \) which seems incomplete. Let's proceed with the following steps: 1. **Factor the first polynomial** \( 3x^2 - 7x - 6 \). 2. **Factor the second polynomial** \( 9x^4 - 13x^2 + 4 \). I'll start by factoring both polynomials. ### Step 1: Factor \( 3x^2 - 7x - 6 \) To factor this quadratic expression, we can look for two numbers that multiply to \( 3 \times (-6) = -18 \) and add to \( -7 \). The numbers that satisfy this are \( -9 \) and \( 2 \). Now we can rewrite the expression: \[ 3x^2 - 9x + 2x - 6 \] Next, we can group the terms: \[ (3x^2 - 9x) + (2x - 6) \] Factoring out the common factors in each group: \[ 3x(x - 3) + 2(x - 3) \] Now, we can factor out the common binomial factor \( (x - 3) \): \[ (3x + 2)(x - 3) \] ### Step 2: Factor \( 9x^4 - 13x^2 + 4 \) This is a quadratic in terms of \( x^2 \). Let's set \( y = x^2 \), so we rewrite the expression as: \[ 9y^2 - 13y + 4 \] Now we need to find two numbers that multiply to \( 9 \times 4 = 36 \) and add to \( -13 \). The numbers that satisfy this are \( -9 \) and \( -4 \). Now we can rewrite the expression: \[ 9y^2 - 9y - 4y + 4 \] Next, we can group the terms: \[ (9y^2 - 9y) + (-4y + 4) \] Factoring out the common factors in each group: \[ 9y(y - 1) - 4(y - 1) \] Now, we can factor out the common binomial factor \( (y - 1) \): \[ (9y - 4)(y - 1) \] Substituting back \( y = x^2 \): \[ (9x^2 - 4)(x^2 - 1) \] The term \( (x^2 - 1) \) can be factored further as \( (x - 1)(x + 1) \). Thus, the complete factorization of \( 9x^4 - 13x^2 + 4 \) is: \[ (9x^2 - 4)(x - 1)(x + 1) \] ### Summary of Results 1. The factorization of \( 3x^2 - 7x - 6 \) is: \[ (3x + 2)(x - 3) \] 2. The factorization of \( 9x^4 - 13x^2 + 4 \) is: \[ (9x^2 - 4)(x - 1)(x + 1) \] If you need further assistance or have additional questions, feel free to ask!