Factoring Polynomials Practice Questions

To factor this expression, first find the greatest common factor (GCF) of both terms. The GCF of \(6x^2\) and \(12x\) is \(6x\).

Divide each term by the GCF:

\(6x^2 \div 6x = x\)

\(12x \div 6x = 2\)

Write the GCF outside the parentheses with the remaining factors inside:

\(6x^2 + 12x = 6x(x + 2)\)

Note that choice A factors out only 6 instead of the full GCF of \(6x\), and choice C factors out only \(2x\). Always factor out the greatest common factor.

To factor a trinomial in the form \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). Here, we need two numbers that multiply to 12 and add to 7:

\(3 \times 4 = 12\:\) and \(\:3 + 4 = 7\)

Therefore:

\(x^2 + 7x + 12 = (x + 3)(x + 4)\)

You can verify by expanding with the FOIL method:

\((x + 3)(x + 4)\)\(\:= x^2 + 4x + 3x + 12\)\(\:= x^2 + 7x + 12\)

This expression is a difference of squares. A difference of squares follows the pattern:

\(a^2 – b^2 = (a + b)(a – b)\)

Here, \(x^2 = (x)^2\) and \(9 = (3)^2\), so:

\(x^2 – 9 = (x + 3)(x – 3)\)

Note that choice A is incorrect because \((x – 3)^2 = x^2 – 6x + 9\), which is not the same as \(x^2 – 9\).

This is a perfect square trinomial. A perfect square trinomial follows the pattern:

\(a^2 – 2ab + b^2 = (a – b)^2\)

Check whether \(x^2 – 10x + 25\) fits this pattern:

• \(a^2 = x^2\), so \(a = x\)
• \(b^2 = 25\), so \(b = 5\)
• \(2ab = 2(x)(5) = 10x\) ✓

Since the middle term matches, we can write:

\(x^2 – 10x + 25 = (x – 5)^2\)

When the leading coefficient is not 1, multiply it by the constant term: \(2 \times 3 = 6\). Find two numbers that multiply to 6 and add to 7:

\(1 \times 6 = 6\:\) and \(\:1 + 6 = 7\)

Rewrite the middle term using these numbers and factor by grouping:

\(2x^2 + 1x + 6x + 3\)

\(= x(2x + 1) + 3(2x + 1)\)

\(= (2x + 1)(x + 3)\)

Find two numbers that multiply to \(-14\) and add to \(-5\). Since the product is negative, the two numbers must have different signs:

\(-7 \times 2 = -14\:\) and \(\:-7 + 2 = -5\)

Therefore:

\(x^2 – 5x – 14 = (x – 7)(x + 2)\)

Be careful with the signs. Choice A reverses them, which would give \(x^2 + 5x – 14\) instead.

This expression is a difference of cubes. The difference of cubes follows the pattern:

\(a^3 – b^3\)\(= (a – b)(a^2 + ab + b^2)\)

Here, \(8x^3 = (2x)^3\) and \(27 = (3)^3\), so \(a = 2x\) and \(b = 3\).

Substitute into the formula:

\(= (2x – 3)((2x)^2 + (2x)(3) + (3)^2)\)

\(= (2x – 3)(4x^2 + 6x + 9)\)

A common mistake is to use a minus sign in the trinomial factor. Remember: for the difference of cubes, the trinomial factor always uses a plus sign on the middle term.

First, factor out the GCF of 3:

\(3x^2 – 12 = 3(x^2 – 4)\)

Now notice that \(x^2 – 4\) is a difference of squares, which can be factored further:

\(x^2 – 4 = (x + 2)(x – 2)\)

Putting it all together:

\(3x^2 – 12 = 3(x + 2)(x – 2)\)

Choice A is partially correct but not completely factored, since \(x^2 – 4\) can still be broken down.

This is a perfect square trinomial. It follows the pattern:

\(a^2 + 2ab + b^2 = (a + b)^2\)

Check whether \(x^2 + 6x + 9\) fits:

• \(a^2 = x^2\), so \(a = x\)
• \(b^2 = 9\), so \(b = 3\)
• \(2ab = 2(x)(3) = 6x\) ✓

Since the middle term matches:

\(x^2 + 6x + 9 = (x + 3)^2\)

This is a difference of squares. Identify the square roots of each term:

\(4x^2 = (2x)^2\) and \(25 = (5)^2\)

Apply the difference of squares formula:

\(a^2 – b^2 = (a + b)(a – b)\)

\(4x^2 – 25 = (2x + 5)(2x – 5)\)

Multiply the leading coefficient by the constant term:

\(6 \times (-10) = -60\)

Find two numbers that multiply to −60 and add to 11:

\(15 \times (-4) = -60\:\) and \(\:15 + (-4) = 11\)

Rewrite the middle term and factor by grouping:

\(6x^2 + 15x – 4x – 10\)

\(= 3x(2x + 5) – 2(2x + 5)\)

\(= (3x – 2)(2x + 5)\)

Watch out for sign errors. Choice B has the correct numbers but the wrong signs, which would give \(6x^2 – 11x – 10\) instead.

First, factor out the GCF. Each term contains at least one factor of \(x\):

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

Now factor the trinomial inside the parentheses. Find two numbers that multiply to 6 and add to 5:

\(2 \times 3 = 6\:\) and \(\:2 + 3 = 5\)

Therefore:

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

To factor by grouping, split the polynomial into two pairs and factor each pair separately:

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

Factor the GCF from each group:

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

Both groups now share the common binomial factor \((x + 3)\). Factor it out:

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

Check whether this is a perfect square trinomial using the pattern \(a^2 – 2ab + b^2 = (a – b)^2\):

• \(a^2 = 9x^2\), so \(a = 3x\)
• \(b^2 = 16\), so \(b = 4\)
• \(2ab = 2(3x)(4) = 24x\) ✓

The middle term matches, so:

\(9x^2 – 24x + 16 = (3x – 4)^2\)

Note that choice C would give \(9x^2 – 16\) (a difference of squares), not the original trinomial.

Recognize that \(x^4 – 16\) is a difference of squares, since \(x^4 = (x^2)^2\) and \(16 = (4)^2\):

\(x^4 – 16 = (x^2 + 4)(x^2 – 4)\)

But we are not done! The factor \(x^2 – 4\) is itself a difference of squares and can be factored further:

\(x^2 – 4 = (x + 2)(x – 2)\)

The factor \(x^2 + 4\) is a sum of squares, which cannot be factored over the real numbers. The completely factored form is:

\(x^4 – 16 = (x^2 + 4)(x + 2)(x – 2)\)

Choice A is partially factored but not complete, since \(x^2 – 4\) can still be broken down.