91Ó°ÊÓ

To factor a polynomial, we first find the Greatest Common Factor (GCF). The GCF is the largest factor that can divide each term of the polynomial evenly. Consider our polynomial:

• \(40 y^{4}\)
• \(4 y^{3}\)
• \(-12 y^{2}\)

First, look at the numerical coefficients (40, 4, and -12). The GCF of these numbers is 4. Next, look at the variable part. Each term has at least \(y^{2}\). Combining these, the GCF of the entire polynomial is \(4y^{2}\). This step helps to simplify the polynomial, making further factoring easier!

In polynomial factorization, we rewrite the polynomial as a product of simpler polynomials. Once we've found the GCF, we factor it out.Here, our polynomial is: \(40 y^{4}+4 y^{3}-12 y^{2}\). We found the GCF to be \(4y^{2}\), so we factor it out:
\(40 y^{4} = 4y^{2} \times 10 y^{2}\)
\(4 y^{3} = 4 y^{2} \times y\)
\(-12 y^{2} = -4 y^{2} \times 3\)
Rewriting the original polynomial, we get:\[ 40 y^{4} + 4 y^{3} - 12 y^{2} = 4y^{2}(10 y^{2} + y - 3) \]This expression shows the polynomial factored with the GCF \(4y^{2}\) outside. Checking further, the quadratic \(10 y^{2} + y - 3\) is not factorable over the integers, so our final factorization is \(4y^{2}(10 y^{2} + y - 3)\).

Quadratic expressions take the form \(ax^2 + bx + c\). In our problem, after factoring out the GCF, we deal with the quadratic \(10 y^{2} + y - 3\). Initially, checking if it can be factored further is essential.Not all quadratics are factorable over integers. One quick check is using the discriminant (\(b^2 - 4ac\)). If the discriminant is not a perfect square, the quadratic is not factorable over the integers.Here, the discriminant for \(10 y^{2} + y - 3\) would be:
\(1^2 - 4 \cdot 10 \cdot (-3) = 1 + 120 = 121\)
121 is a perfect square, suggesting potential factorability. However, factor pairs analysis of \(10\) and \(-3\) that satisfy the equation fails, confirming non-integral factors. Hence, \(10 y^{2} + y - 3\) remains.By understanding quadratics, it’s easier to judge when further factoring is impossible.