Factoring Polynomials into Linear Factors — Practice Questions

Below are 8 practice questions on factoring polynomials into products of completely linear factors. The problems cover the difference of squares, factor by grouping, the Rational Root Theorem, synthetic division, complex roots, and zero verification. Complete step-by-step solutions are provided beneath each question — ideal for algebra and precalculus students.

Question 1 — Difference of Squares

Write \( P(x) = x^2 - 4 \) as a product of linear factors.

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Apply \(a^2 - b^2 = (a-b)(a+b)\) with \(a = x\), \(b = 2\).

Answer: \( P(x) = (x - 2)(x + 2) \)

Question 2 — Rational Root Theorem and Quadratic Factoring

Write \( P(x) = 9x^3 - 19x - 10 \) as a product of linear factors.

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Test \(x = -1\):

\[ P(-1) = -9 + 19 - 10 = 0 \]

So \((x+1)\) is a factor. Dividing gives:

\[ 9x^2 - 9x - 10 \]

Factor: two numbers multiply to \(-90\) and add to \(-9\) → \((-15)\) and \(6\).

\[ 9x^2 - 15x + 6x - 10 = 3x(3x - 5) + 2(3x - 5) = (3x+2)(3x-5) \]

Answer: \( P(x) = (x + 1)(3x + 2)(3x - 5) \)

Question 3 — Factor by Grouping with Complex Roots

Write \( P(x) = 5x^3 - 2x^2 + 5x - 2 \) as a product of linear factors.

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Group and factor:

\[ (5x^3 - 2x^2) + (5x - 2) = x^2(5x-2) + 1(5x-2) = (x^2+1)(5x-2) \]

Factor \(x^2 + 1\) over \(\mathbb{C}\): \(x = \pm i\), so \(x^2 + 1 = (x-i)(x+i)\).

Answer: \( P(x) = (x + i)(x - i)(5x - 2) \)

Question 4 — Negative Leading Coefficient

Write \( P(x) = -x^3 - 2x^2 + 5x + 6 \) as a product of linear factors.

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Test \(x = -1\):

\[ P(-1) = 1 - 2 - 5 + 6 = 0 \]

So \((x+1)\) is a factor. Dividing gives \(-x^2 - x + 6 = -(x^2 + x - 6) = -(x+3)(x-2)\).

Putting it together:

\[ P(x) = -1(x+1)(x+3)(x-2) \]

Equivalently, absorbing the \(-1\) into the last factor, \( (x+1)(x+3)(2-x) \).

Answer: \( P(x) = -1(x + 1)(x - 2)(x + 3) \), or equivalently \((x + 1)(2 - x)(x + 3)\)

Question 5 — Strictly Monic Linear Factors

Write \( P(x) = (x + 2)(20x^2 + 41x + 20) \) as a product of strictly linear factors of the form \((x - c)\).

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Factor the quadratic. Two numbers multiply to \(400\) and add to \(41\) → \(16\) and \(25\):

\[ 20x^2 + 16x + 25x + 20 = 4x(5x+4) + 5(5x+4) = (4x+5)(5x+4) \]

Now write each binomial with leading coefficient \(1\) by pulling out the constant:

\[ 4x + 5 = 4\!\left(x + \tfrac{5}{4}\right), \quad 5x + 4 = 5\!\left(x + \tfrac{4}{5}\right) \]

The product \(4 \cdot 5 = 20\) appears as the overall leading coefficient.

Answer: \( P(x) = 20(x + 2)\!\left(x + \tfrac{5}{4}\right)\!\left(x + \tfrac{4}{5}\right) \)

Question 6 — Verify Two Real Zeros, Find the Rest

Show that \(1\) and \(-3\) are zeros of \( P(x) = x^4 + 4x^3 + 6x^2 + 4x - 15 \), then express \(P(x)\) as a product of linear factors.

View Solution

Evaluate:

\[ P(1) = 1 + 4 + 6 + 4 - 15 = 0 \] \[ P(-3) = 81 - 108 + 54 - 12 - 15 = 0 \]

So \((x-1)\) and \((x+3)\) are factors; their product is \(x^2 + 2x - 3\).

Divide:

\[ \frac{x^4 + 4x^3 + 6x^2 + 4x - 15}{x^2 + 2x - 3} = x^2 + 2x + 5 \]

Quadratic formula on \(x^2 + 2x + 5 = 0\):

\[ x = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i \]

Answer: \( P(x) = (x - 1)(x + 3)(x + 1 - 2i)(x + 1 + 2i) \)

Question 7 — Two Negative Real Zeros

Show that \(-1\) and \(-5\) are zeros of \( P(x) = x^4 + 12x^3 + 54x^2 + 108x + 65 \), then express \(P(x)\) as a product of linear factors.

View Solution

Evaluate:

\[ P(-1) = 1 - 12 + 54 - 108 + 65 = 0 \] \[ P(-5) = 625 - 1500 + 1350 - 540 + 65 = 0 \]

The factors \((x+1)\) and \((x+5)\) have product \(x^2 + 6x + 5\). Divide:

\[ \frac{x^4 + 12x^3 + 54x^2 + 108x + 65}{x^2 + 6x + 5} = x^2 + 6x + 13 \]

Quadratic formula on \(x^2 + 6x + 13 = 0\):

\[ x = \frac{-6 \pm \sqrt{-16}}{2} = -3 \pm 2i \]

Answer: \( P(x) = (x + 1)(x + 5)(x + 3 - 2i)(x + 3 + 2i) \)

Question 8 — Three Real Zeros in a Degree-5 Polynomial

Show that \(-2\), \(-3\), and \(2\) are zeros of \( P(x) = x^5 + x^4 + 26x^2 - 16x - 120 \), then write \(P(x)\) as a product of linear factors.

View Solution

Evaluate (note the missing \(x^3\) term, coefficient \(0\)):

\[ P(-2) = -32 + 16 + 104 + 32 - 120 = 0 \] \[ P(-3) = -243 + 81 + 234 + 48 - 120 = 0 \] \[ P(2) = 32 + 16 + 104 - 32 - 120 = 0 \]

The corresponding factors \((x+2)\), \((x+3)\), \((x-2)\) multiply to:

\[ (x^2 - 4)(x + 3) = x^3 + 3x^2 - 4x - 12 \]

Divide:

\[ \frac{x^5 + x^4 + 26x^2 - 16x - 120}{x^3 + 3x^2 - 4x - 12} = x^2 - 2x + 10 \]

Quadratic formula on \(x^2 - 2x + 10 = 0\):

\[ x = \frac{2 \pm \sqrt{-36}}{2} = 1 \pm 3i \]

Answer: \( P(x) = (x + 2)(x + 3)(x - 2)(x - 1 - 3i)(x - 1 + 3i) \)

Frequently Asked Questions

What does "product of linear factors" mean?

It means writing the polynomial as a product of the form \(a(x - c_1)(x - c_2)\cdots(x - c_n)\), where each \(c_k\) is a root (possibly complex) and \(a\) is the leading coefficient.

What is factor by grouping?

When a polynomial has four terms, group them into two pairs, factor a GCF from each pair, then factor out the shared binomial. Used in Questions 2 and 3.

How do complex roots appear in linear factorization?

An irreducible quadratic like \(x^2 + 1\) factors into \((x - i)(x + i)\). For real-coefficient polynomials, complex roots always appear in conjugate pairs \(a \pm bi\).

How can I verify my factorization?

Multiply the factors back together and check that you recover the original polynomial, or substitute any root \(c\) into the original and confirm \(P(c) = 0\).