Practice Factoring Trinomials

To factor ‍\(x^2+7x+{10}\), we need to find factors of ‍ 10 that add up to 7.

The table below shows all possible factors of 10 and their sums.

Only one pair of factors satisfies the conditions of the problem: ‍2 and 5.

We can add each of these numbers to \(x\) to form the two binomial factors: ‍\((x+2)\) and ‍\((x+5)\). The factorization of the polynomial is given below.

\(x^2+7x+10=(x+2)(x+5)\)

To factor ‍\(x^2+9x+{20}\), we need to find factors of ‍ 20 that add up to 9.

The table below shows all possible factors of 20 and their sums.

Only one pair of factors satisfies the conditions of the problem: ‍4 and 5.

We can add each of these numbers to \(x\) to form the two binomial factors: ‍\((x+4)\) and ‍\((x+5)\). The factorization of the polynomial is given below.

\(x^2+9x+20=(x+4)(x+5)\)

To factor ‍\(x^2-8x-9\), we need to find factors of ‍-9 that add up to -8.

Since the two numbers must multiply to -9, one must be positive and one must be negative.

The factors of ‍-9 are listed in the table below.

Notice here that ‍1 and ‍-9 satisfy these requirements. We can add each of these numbers to ‍\(x\) to form the two binomial factors: ‍\((x+1)\) and ‍\((x+(-9))\). The factorization of the polynomial is given below.

\( x^2-8x-9 =(x+1)(x+({-9})) \\ \quad =(x+1)(x-9) \)

To factor ‍\(x^2-10x+24\), we need to find factors of ‍24 that add up to -8.

Since the two numbers must multiply to 24 and add to -10, both numbers must be negative.

The factors of ‍24 (where both are negative) are listed in the table below.

Notice here that ‍-4 and ‍-6 satisfy these requirements. We can add each of these numbers to ‍\(x\) to form the two binomial factors: ‍\((x+(-4))\) and ‍\((x+(-6))\). The factorization of the polynomial is given below.

\( x^2-10x-9 =(x+(-4))(x+({-6})) \\\quad =(x-4)(x-6) \)

\(=(x-3)(x+10)\)

No

In this method, we factor the trinomial by equating it to ‍\((x+m)(x+n)\) for some integers ‍\(m\) and ‍\(n\). This means that the leading term of this polynomial will always be ‍\(x^2\). Since the leading term of this polynomial is ‍\(2x^2\), it cannot be factored using this method.

Recall the following factorization:

\(x^2+5x+6=(x+2)(x+3)\)

The polynomial ‍\(x^2+5xy+6y^2\) is very similar to ‍\(x^2+5x+6\). The only difference is that the middle term is multiplied by ‍\(y\) and the last term is multiplied by ‍\(y^2\).

We can obtain this by adding a factor of ‍\(y\) to each binomial:

\(\begin{aligned}(x+2y)(x+3y)
&=x^2+3xy +2x y+6{y^2}\\
\\
&=x^2+5xy+6{y^2}
\end{aligned}\)

The factorization is \((x+2y)(x+3y)\)

This polynomial is of the form ‍\(x^2+bx+c\), although a substitution makes this easier to see.

Let \(Y=x^2\). We can now write the polynomial as follows:

\(\begin{aligned}x^4-5x^2+6&=({x^2})^2-5{x^2}+6\\
\\
&={Y}^2-5{Y}+6\\
\end{aligned}\)

Since \((-2)\cdot (-3)=6\) and \((-2)+(-3)=-5\), the polynomial factors as follows:

\(=(Y-2)(Y-3)\)

Now, since \(Y=x^2\), we can back substitute to find a factorization of the original polynomial.

\((x^2-2)(x^2-3)\)

The factorization is \((x^2-2)(x^2-3)\).