Zeros and Linear Factors
A polynomial can be written in several forms. Its expanded form shows its terms and coefficients, while a factored form displays products that may reveal important inputs. In particular, a factor such as \(x-3\) makes the polynomial equal to zero when \(x=3\).
This connection is useful when interpreting polynomial equations. A zero is an input at which the polynomial's value is zero; a matching linear factor is an algebraic way to record that input. We will translate between these descriptions and check the results by substitution and multiplication.
The expression \(x-r\) is the linear factor associated with \(r\). The key sign detail is that setting \(x-r=0\) gives \(x=r\), not \(x=-r\). Thus, a factor \(x+2\), written as \(x-(-2)\), corresponds to the input \(-2\).
Why a Linear Factor Gives a Zero
Suppose a polynomial is written as \(P(x)=(x-r)Q(x)\), where \(Q\) is another polynomial. Substituting \(x=r\) makes the first factor zero. The product is therefore zero, whatever value \(Q(r)\) has.
This is a complete justification of the direction from factor to zero: if \(x-r\) is a factor, then \(r\) is a zero. In the other direction, a polynomial that has zero \(r\) can be expressed with \(x-r\) as a factor. The examples below show that relationship in practice; the general algebraic justification for the converse requires a separate argument.
Worked Examples
Worked Example: Read the Zeros and Expand a Factored Polynomial
Consider \(P(x)=2(x-3)(x+1)\). First, use the factors to identify the inputs that make the product zero. The factor \(x-3\) is zero at \(x=3\), and \(x+1=x-(-1)\) is zero at \(x=-1\).
To translate the expression into expanded form, multiply the binomials:
Now multiply by the outside coefficient:
Thus the corresponding polynomial equation can be written in either form:
Check both candidate zeros in the expanded polynomial. At \(x=3\), \(P(3)=2(3)^2-4(3)-6=18-12-6=0\). At \(x=-1\), \(P(-1)=2(-1)^2-4(-1)-6=2+4-6=0\). Both substitutions confirm the zeros read from the factors.
Worked Example: Rewrite a Polynomial Equation in Factored Form
Consider \(x^2-7x+12=0\). We seek two numbers whose product is \(12\) and whose sum is \(-7\). The numbers \(-3\) and \(-4\) have those properties, so the trinomial factors as \((x-3)(x-4)\): the cross terms sum to \(-7x\), and the constant product is \(12\).
Therefore, the equation has the equivalent factored form \((x-3)(x-4)=0\). The factors indicate the inputs \(x=3\) and \(x=4\). To check, substitute each into the original polynomial:
The factored form is confirmed by expansion, and both indicated inputs make the original polynomial zero.
Worked Example: Track a Repeated Linear Factor
Let \(P(x)=-3(x+2)^2(x-1)\). The factor \(x+2=x-(-2)\) gives the zero \(-2\), and \(x-1\) gives the zero \(1\). The factor \((x+2)^2\) shows that the factor for \(-2\) occurs twice. For now, the important point is that it still identifies the same input, \(-2\).
Expand to obtain a polynomial equation in standard form. First square the binomial:
Then multiply by \(x-1\):
Finally, multiply each term by \(-3\):
Check the zeros in the expanded form. For \(x=-2\),
For \(x=1\),
Both checks agree with the factors in the original expression.
Worked Example: Check a Zero and Its Linear Factor
Consider \(P(x)=x^3-2x^2-5x+6\). Substituting \(x=1\) gives
So \(1\) is a zero. A factorization that displays the corresponding linear factor is \((x-1)(x-3)(x+2)\). Verify it by multiplying:
and then
The product matches \(P(x)\), and its factor \(x-1\) corresponds to the zero \(1\). The other displayed factors identify \(-2\) and \(3\). As a direct check, \(P(-2)=-8-8+10+6=0\), and \(P(3)=27-18-15+6=0\).
Translating Between the Two Forms
A polynomial equation is an equation in which a polynomial expression is set equal to a value, often zero. When \(P(x)=0\), its factored form makes the relevant zero candidates visible. Conversely, expanding the factors gives an equivalent equation in polynomial standard form. Each translation must preserve equality: expand every factor, retain any outside coefficient, and keep the equation's right-hand side unchanged unless the same operation is applied to both sides.
For each factor \(x-r\), record the associated input \(r\). A factor \(x+a\) is \(x-(-a)\), so its associated input is \(-a\).
Multiply the factors carefully, then include any constant coefficient multiplying the product.
Substitute each recorded input into the polynomial and confirm that the value is zero. If you began with an expanded polynomial, multiply the proposed factors back out to confirm the identity.
An outside nonzero constant does not change which inputs make a product zero. For example, \(5(x-2)(x+4)=0\) has the same zeros as \((x-2)(x+4)=0\), since \(5\) is not zero. This observation would not allow division by a zero coefficient: the nonzero condition matters.
Common Sign Mistake
A frequent error is to read the number in a factor with the wrong sign. For example, from \(x+4\), it is incorrect to claim that the associated input is \(4\). Setting the factor to zero shows why:
Indeed, if \(P(x)=(x+4)(x-2)\), then \(P(-4)=(-4+4)(-4-2)=0\). But \(P(4)=(4+4)(4-2)=8\cdot2=16\), not zero. Rewrite \(x+4\) as \(x-(-4)\) to keep the sign clear.
Why the Connection Matters
The expanded form is useful for reading the polynomial's terms, while the factored form can reveal its zeros directly. Moving between them lets you check that a proposed factorization represents the same polynomial and understand what its factors say about inputs. Substitution is an essential check: a proposed zero \(r\) is valid only if evaluating the original polynomial gives \(P(r)=0\).
Keep the central relationship precise: a factor \(x-r\) guarantees that \(r\) is a zero, because substitution makes that factor vanish. When a zero is known, the polynomial can be rewritten to expose the matching linear factor; multiplication verifies such a factorization in any particular example.
Check Your Understanding
Use substitution or multiplication to justify your answers.
- If \(P(x)=(x-5)(x+2)\), which inputs are indicated as zeros by the factors?
- Expand \((x-1)(x+3)\) and write the equivalent equation with the expanded polynomial on the left.
- For \(P(x)=4(x+6)(x-2)\), explain why the outside coefficient does not change the zeros.
- What sign error could occur when reading a zero from \(x+7\), and what is the correct input?
- For \(P(x)=(x-3)^2(x+1)\), which input is associated with the repeated factor?
