Factor $-20h^2 - 11h + 3$ Completely

When we talk about factoring polynomials, we're essentially trying to break them down into simpler expressions that, when multiplied together, give us the original polynomial. It's like finding the prime factors of a number, but for algebraic expressions. The expression we're working with today is a quadratic trinomial: $-20h^2 - 11h + 3$. Our goal is to find two binomials that multiply to give us this exact expression. This process can seem a bit daunting at first, especially with the negative leading coefficient and the specific middle term, but with a systematic approach, it becomes quite manageable. We'll explore different methods to tackle this, ensuring we cover all the bases to achieve a complete factorization. Remember, a polynomial is factored completely when its factors cannot be factored any further. This means each factor is either a prime polynomial or a monomial.

Let's begin by addressing the negative leading coefficient. It's often easier to work with quadratic expressions when the coefficient of the squared term is positive. So, our first strategic move is to factor out a $-1$ from the entire expression. This transforms $-20h^2 - 11h + 3$ into $-1(20h^2 + 11h - 3)$. Now, we can focus on factoring the trinomial inside the parentheses: $20h^2 + 11h - 3$. This is a standard form of a quadratic trinomial $ax^2 + bx + c$, where $a=20$, $b=11$, and $c=-3$. Our mission is to find two binomials of the form $(ph + q)(rh + s)$ such that when expanded, they result in $20h^2 + 11h - 3$. The product of the first terms ($ph imes rh$) must equal $20h^2$, and the product of the last terms ($q imes s$) must equal $-3$. Additionally, the sum of the outer and inner products ($phs + qrh$) must equal $11h$, which is our middle term.

One popular method for factoring trinomials of the form $ax^2 + bx + c$ is the ac method. Here, we first calculate the product of $a$ and $c$. In our case, $a=20$ and $c=-3$, so $ac = 20 imes (-3) = -60$. Now, we need to find two numbers that multiply to $-60$ and add up to $b$, which is $11$. Let's list pairs of factors for $-60$ and check their sums:

• 1 and -60 (sum: -59)
• -1 and 60 (sum: 59)
• 2 and -30 (sum: -28)
• -2 and 30 (sum: 28)
• 3 and -20 (sum: -17)
• -3 and 20 (sum: 17)
• 4 and -15 (sum: -11)
• -4 and 15 (sum: 11)

Success! The pair of numbers that multiply to $-60$ and add up to $11$ are $-4$ and $15$. These numbers will help us rewrite the middle term ($11h$) as the sum of two terms. So, $20h^2 + 11h - 3$ can be rewritten as $20h^2 - 4h + 15h - 3$.

The next step in the ac method is to use grouping. We group the first two terms and the last two terms: $(20h^2 - 4h) + (15h - 3)$. Now, we factor out the greatest common factor (GCF) from each group. In the first group, $20h^2 - 4h$, the GCF is $4h$. Factoring this out, we get $4h(5h - 1)$. In the second group, $15h - 3$, the GCF is $3$. Factoring this out, we get $3(5h - 1)$. So, our expression now looks like $4h(5h - 1) + 3(5h - 1)$.

Notice that both terms now share a common binomial factor, $(5h - 1)$. This is a crucial sign that we are on the right track! We can now factor out this common binomial: $(5h - 1)(4h + 3)$. This is the factored form of $20h^2 + 11h - 3$. Since we initially factored out a $-1$ from the original expression, we need to reintroduce it to get the complete factorization of $-20h^2 - 11h + 3$. Therefore, the complete factorization is $-1(5h - 1)(4h + 3)$.

We can also express this by distributing the negative sign to one of the binomial factors. For example, distributing the $-1$ to $(5h - 1)$ gives us $(-5h + 1)(4h + 3)$. Alternatively, distributing the $-1$ to $(4h + 3)$ gives us $(5h - 1)(-4h - 3)$. All these forms are considered correct factorizations. It's always a good idea to verify your answer by multiplying the factors back together to ensure you get the original polynomial. Let's check: $-1(5h - 1)(4h + 3) = -1(20h^2 + 15h - 4h - 3) = -1(20h^2 + 11h - 3) = -20h^2 - 11h + 3$. The verification confirms our factorization is correct.

Another method to consider is the trial and error method, especially for trinomials where the numbers involved aren't too large. For $20h^2 + 11h - 3$, we need to find two binomials $(ph + q)(rh + s)$. We know that $p imes r$ must equal $20$, and $q imes s$ must equal $-3$. The pairs for $p$ and $r$ could be (1, 20), (2, 10), (4, 5). The pairs for $q$ and $s$ could be (1, -3) or (-1, 3). We need to systematically try combinations and check if the sum of the outer and inner products equals $11h$. For example, let's try $(4h + q)(5h + s)$.

If we choose $q=1$ and $s=-3$, we get $(4h + 1)(5h - 3)$. Expanding this: $20h^2 - 12h + 5h - 3 = 20h^2 - 7h - 3$. This is not our target. Let's try switching the signs of $q$ and $s$: $(4h - 1)(5h + 3)$. Expanding this: $20h^2 + 12h - 5h - 3 = 20h^2 + 7h - 3$. Still not there.

What if we switch the order of the factors for 20? Let's try $(5h + q)(4h + s)$. Let's use $q=-1$ and $s=3$: $(5h - 1)(4h + 3)$. Expanding this: $20h^2 + 15h - 4h - 3 = 20h^2 + 11h - 3$. Bingo! This matches our trinomial. So, $(5h - 1)(4h + 3)$ is the factored form of $20h^2 + 11h - 3$. Again, remembering the initial $-1$ we factored out, the complete factorization of $-20h^2 - 11h + 3$ is $-1(5h - 1)(4h + 3)$.

It's important to note that factoring is a fundamental skill in algebra and has numerous applications, from solving quadratic equations to simplifying complex algebraic expressions and analyzing functions. The ability to factor polynomials efficiently can significantly simplify problem-solving. For instance, when solving an equation like $-20h^2 - 11h + 3 = 0$, factoring it into $-1(5h - 1)(4h + 3) = 0$ immediately tells us that either $5h - 1 = 0$ or $4h + 3 = 0$. Solving these simple linear equations gives us $h = 1/5$ and $h = -3/4$, which are the roots of the original quadratic equation. This demonstrates the power and utility of complete factorization.

Remember, practice is key! The more you practice factoring different types of polynomials, the more comfortable and proficient you will become. Don't be discouraged if you don't get it right on the first try. Every attempt, successful or not, is a learning opportunity. Explore various examples, use different methods if one isn't clicking, and always, always check your work. Mastering factorization is a significant step in building a strong foundation in mathematics.

For further exploration into algebraic concepts and factoring techniques, you can visit Khan Academy, a fantastic resource for learning mathematics online.