Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ y^{4}+11 y^{3}+30 y^{2} $$

Short answer

The factored form is \(y^2(y + 5)(y + 6)\).

Step-by-step solution

Identify Common Factors

Look for any common factors in each term of the polynomial. The polynomial is \(y^4 + 11y^3 + 30y^2\). All terms have at least a \(y^2\) in common, so factor \(y^2\) out of the polynomial.

Factor out the Common Factor

Factor \(y^2\) out of each term: \(y^2(y^2 + 11y + 30)\).

Factor the Quadratic Polynomial

Now, focus on factoring the quadratic polynomial \(y^2 + 11y + 30\). Find two numbers that multiply to 30 and add up to 11. These numbers are 5 and 6.

Write the Quadratic as a Product of Binomials

Write \(y^2 + 11y + 30\) as a product of two binomials: \((y + 5)(y + 6)\).

Combine All Factors

Combine \(y^2\) with the factored quadratic: \(y^2(y + 5)(y + 6)\). This is the completely factored form of the polynomial.

Key concepts

common factors

To factor a polynomial completely, the first and most crucial step is to identify any common factors present in each term. This means searching for a variable or number that is a factor of every term in the polynomial. In the given polynomial \( y^{4} + 11y^{3} + 30y^{2} \) , we notice that each term contains at least \( y^{2} \). Factoring out the greatest common factor (GCF) simplifies the polynomial and is a necessary step before proceeding further with factoring techniques. By factoring out common factors, you reduce the polynomial to a simpler form, making the subsequent steps more straightforward. For instance, factoring \( y^{2} \) from the given polynomial yields: \( y^{2}(y^{2} + 11y + 30) \).

quadratic polynomial

The next focus is on the quadratic polynomial left inside the parentheses, \( y^{2} + 11y + 30 \). Quadratic polynomials are second-degree polynomials typically of the form \( ax^{2} + bx + c \). Factoring quadratics often involves finding two numbers whose product is equal to the constant term \( c \), and whose sum is equal to the linear coefficient \( b \). This particular quadratic polynomial can be expressed as a product of two binomials.

binomials

Binomials are polynomials with exactly two terms. In our example, once we identify the numbers that multiply to \( 30 \) (the constant term) and add up to \( 11 \) (the linear coefficient), we can rewrite the quadratic as a product of two binomials. These numbers are \( 5 \) and \( 6 \) because \( 5 \times 6 = 30 \) and \( 5 + 6 = 11 \). Therefore, we can factor the quadratic polynomial as follows: \( y^{2} + 11y + 30 = (y + 5)(y + 6) \). Understanding this step is crucial, as it transforms the quadratic into a simpler, more manageable form.

factoring techniques

There are several techniques for factoring polynomials, but in this exercise, we mainly use common factors and factoring quadratics using the sum and product method. To summarize the process:

• First, we find and factor out the \( GCF \); here, it’s \( y^{2} \).
• Next, we work on factoring the quadratic polynomial \( y^{2} + 11y + 30 \) by finding two numbers that multiply to \( 30 \) and add up to \( 11 \). These numbers are \( 5 \) and \( 6 \).
• We then express the quadratic as a product of two binomials:\( (y + 5)(y + 6) \).
• Finally, we combine the common factor with the factored quadratic to get the completely factored form: \( y^{2}(y + 5)(y + 6) \).

Knowing these techniques can help you factor even more complex polynomials. Practice is key to mastering these steps!