Question

1. $-14u^{4}v^{3} + 6u^{2}v^{4} - 10u^{2}v^{2}$

Explanation:

Step1: Identify the GCF of coefficients and variables

The coefficients are -14, 6, -10. The GCF of 14, 6, 10 is 2 (and we take -2 to keep the leading coefficient positive? Wait, actually GCF of -14,6,-10 is -2? Wait no, GCF of 14,6,10 is 2. Let's check: 14=2×7, 6=2×3, 10=2×5. So GCF of coefficients is 2 (but since the first term is negative, we can factor out -2 or 2. Let's see the variables: each term has \(u^2\) (since \(u^4 = u^2 \times u^2\), \(u^2 = u^2 \times 1\)) and \(v^2\) (since \(v^3 = v^2 \times v\), \(v^4 = v^2 \times v^2\), \(v^2 = v^2 \times 1\)). So the GCF of the terms is \(-2u^2v^2\)? Wait, let's factor out \(-2u^2v^2\):

\(-14u^4v^3 = -2u^2v^2 \times 7u^2v\)

\(6u^2v^4 = -2u^2v^2 \times (-3v^2)\)

\(-10u^2v^2 = -2u^2v^2 \times 5\)

Wait, maybe better to factor out \( -2u^2v^2 \):

Step2: Factor out the GCF

Factor out \(-2u^2v^2\) from each term:

\(-14u^4v^3 + 6u^2v^4 - 10u^2v^2 = -2u^2v^2(7u^2v - 3v^2 + 5)\)

Wait, let's check:

\(-2u^2v^2 \times 7u^2v = -14u^4v^3\)

\(-2u^2v^2 \times (-3v^2) = 6u^2v^4\)

\(-2u^2v^2 \times 5 = -10u^2v^2\)

Yes, that works. So the factored form is \(-2u^2v^2(7u^2v - 3v^2 + 5)\)

Alternatively, we can factor out \(2u^2v^2\) with a negative sign:

Wait, another way: the GCF of the coefficients (ignoring sign) is 2, and the variables: the lowest power of u is \(u^2\), lowest power of v is \(v^2\). So GCF is \(2u^2v^2\), but since the first term is negative, factoring out \(-2u^2v^2\) makes the first term in the parentheses positive.

So:

\(-14u^4v^3 + 6u^2v^4 - 10u^2v^2 = -2u^2v^2(7u^2v - 3v^2 + 5)\)

Answer:

\(-2u^2v^2(7u^2v - 3v^2 + 5)\)