Q&A for How to Factor Trinomials

The constant is +pq.

Yes, the term is actually "slip and slide." The explanation is too lengthy to treat here, but you can websearch it.

Yes. Since you are multiplying the exact same trionimals the answer will be the same.

By inspecting the coefficient of the first term and the third-term constant. The factors of those terms must be combinable by multiplication and addition to produce the second-term coefficient.

A cubic, if it factors at all, must have a linear factor and setting that linear factor to zero gives a rational root. Use the rational root theorem to search for rational roots. If it has one, factor it out. If it doesn't, it is already irreducible.

This expression does not factor easily, so you would have to use the quadratic formula: x = [-b +/-√(b² - 4ac)] ÷ 2a.

It is not factorable.

Although that expression is not an equation, it can still be factored this way: first factor (-1) out of (q - p) in the middle term. That changes (q - p) to (p - q) and also changes -17y to +17y. Now the expression reads 3x(p - q) +17y(p - q) + (p - q). Note that there is an "understood" "+1" in front of the final (p - q). There is now a common factor (p - q) in all three terms of the trinomial. Factor out that common factor, and you're done: (p - q)(3x + 17y + 1).

21x² + 11x - 2 = (7x - 1)(3x+ 2).

Not necessarily. For example (x^2+1)(x+1) = x^3 +x^2 +x +1 is not a trinomial because it has four terms. However, if both binomials have the same variables to the same powers, then it is true. In general, multiplying binomials gives four terms, one corresponding to each letter of the FOIL acronym. So, you only get a trinomial when the O and I terms combine.

That expression does not factor. If that expression were set equal to zero, you would then have a quadratic equation, the roots of which could be found using the quadratic formula.

k² - k - 20 = (k - 5)(k + 4).

Right, it isn't. However, in this example, the distinction between primes and units isn't that important. That's because all the author really needs to show is that the only way do factor 3 as a product of whole numbers is 1x3 and the only way to factor 1 as a product of whole numbers is 1x1.

You can't solve it, because it's not an equation. However, the expression can be factored as x(a + b + 2c).

That expression cannot be factored in any conventional sense.

Notice that the middle term is (-1xy). That means you need two factors of 56 that are just 1 away from each other. 7 and 8 are factors of 56, so use them. y² - xy - 56x² = (y + 7x)(y - 8x).

You're looking for factors of 6 and factors of -2 that will combine (by adding or subtracting) to make 1. 6 will factor into 6 and 1 or 3 and 2. -2 will factor into -2 and +1 or +2 and -1. Choose 3 and 2 for the factors of 6, and +2 and -1 for the factors of -2. Thus, 6z² + z - 2 = (3z + 2)(2z - 1).

A simple trinomial has a coefficient of the "squared" term equal to 1. An example is x² - 3x + 14. A complex trinomial has a coefficient of the "squared" term that is not equal to 1. An example is 2x² - 3x + 14.

That expression won't factor, as you've written it. If there were a positive sign in front of the 7, the expression would factor into (3c + 1)(2c + 7).

Look for two numbers that multiply together to get -12 and add together to get -4. Those numbers are -6 and +2. So the factors are (x - 6) and ( x + 2). Thus, x² - 4x - 12 = (x - 6)(x + 2).

See Factor a Cubic Polynomial .