Question

factor completely: $2a^4 + 4a^3 - 6a^2$
$\big(2a^3 + 6a^2\big)(a - 1)$
$(a + 1)\big(2a^3 - 6a^2\big)$
$2a^2\big(a^2 + 2a - 3\big)$
$2a^2(a + 3)(a - 1)$
question 4
divide: $\frac{6x^2 + 5x - 4}{3x + 4}$
$2x + 5x - 2$
$2x + 3$
$2x - 1$
$2x - 1$

Explanation:

First Problem: Factor completely $2a^4 + 4a^3 - 6a^2$

Step1: Factor out GCF

Identify the greatest common factor (GCF) of all terms, which is $2a^2$.
$2a^4 + 4a^3 - 6a^2 = 2a^2(a^2 + 2a - 3)$

Step2: Factor quadratic trinomial

Factor the quadratic $a^2 + 2a - 3$ into two binomials by finding two numbers that multiply to $-3$ and add to $2$ (these numbers are $3$ and $-1$).
$a^2 + 2a - 3 = (a + 3)(a - 1)$

Step3: Combine factors

Substitute the factored quadratic back into the expression.
$2a^2(a^2 + 2a - 3) = 2a^2(a + 3)(a - 1)$

Step1: Factor numerator

Factor the quadratic numerator $6x^2 + 5x - 4$. Find two numbers that multiply to $6\times(-4)=-24$ and add to $5$ (these numbers are $8$ and $-3$), then rewrite and factor by grouping:
$6x^2 + 8x - 3x - 4 = 2x(3x + 4) -1(3x + 4) = (2x - 1)(3x + 4)$

Step2: Cancel common factors

Cancel the common factor $(3x + 4)$ from numerator and denominator (where $3x + 4
eq 0$).
$\frac{(2x - 1)(3x + 4)}{3x + 4} = 2x - 1$

Answer:

$2a^2 (a + 3) (a - 1)$

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Second Problem: Divide $\frac{6x^2 + 5x - 4}{3x + 4}$