Question

1. $12x^3 + 39x^2 - 36x$

9.

1. $2x^2 - 6xy - 8y^2$

12.

1. $6x^2 - 19xy + 15y^2$

15.

Explanation:

Let's solve each problem one by one:

Problem 8: \(12x^3 + 39x^2 - 36x\)

Step 1: Factor out the GCF

First, find the greatest common factor (GCF) of the terms \(12x^3\), \(39x^2\), and \(-36x\). The GCF of 12, 39, and 36 is 3, and the GCF of \(x^3\), \(x^2\), and \(x\) is \(x\). So we factor out \(3x\):
\(12x^3 + 39x^2 - 36x = 3x(4x^2 + 13x - 12)\)

Step 2: Factor the quadratic

Now, factor the quadratic \(4x^2 + 13x - 12\). We need two numbers that multiply to \(4 \times (-12) = -48\) and add up to 13. The numbers are 16 and -3.
Rewrite the middle term:
\(4x^2 + 16x - 3x - 12\)
Group the terms:
\((4x^2 + 16x) + (-3x - 12)\)
Factor out the GCF from each group:
\(4x(x + 4) - 3(x + 4)\)
Factor out \((x + 4)\):
\((4x - 3)(x + 4)\)

So the fully factored form is:
\(3x(4x - 3)(x + 4)\)

Problem 11: \(2x^2 - 6xy - 8y^2\)

Step 1: Factor out the GCF

The GCF of 2, -6, and -8 is 2. Factor out 2:
\(2x^2 - 6xy - 8y^2 = 2(x^2 - 3xy - 4y^2)\)

Step 2: Factor the quadratic

Factor the quadratic \(x^2 - 3xy - 4y^2\). We need two numbers that multiply to \(-4y^2\) and add up to \(-3y\). The numbers are \(-4y\) and \(y\).
So,
\(x^2 - 3xy - 4y^2 = (x - 4y)(x + y)\)

Thus, the fully factored form is:
\(2(x - 4y)(x + y)\)

Problem 14: \(6x^2 - 19xy + 15y^2\)

Step 1: Factor the quadratic

We need to factor \(6x^2 - 19xy + 15y^2\). We look for two numbers that multiply to \(6 \times 15 = 90\) and add up to \(-19\) (since the middle term is \(-19xy\) and the last term is positive, both numbers are negative). The numbers are \(-10\) and \(-9\).
Rewrite the middle term:
\(6x^2 - 10xy - 9xy + 15y^2\)
Group the terms:
\((6x^2 - 10xy) + (-9xy + 15y^2)\)
Factor out the GCF from each group:
\(2x(3x - 5y) - 3y(3x - 5y)\)
Factor out \((3x - 5y)\):
\((2x - 3y)(3x - 5y)\)

So the fully factored form is:
\((2x - 3y)(3x - 5y)\)

Answer:

s:

1. \(\boldsymbol{3x(4x - 3)(x + 4)}\)
2. \(\boldsymbol{2(x - 4y)(x + y)}\)
3. \(\boldsymbol{(2x - 3y)(3x - 5y)}\)