For the polynomial below, -1 is a zero.

Express as a product of linear factors.

Ask by Lawson Adkins. in the United States

Mar 29,2025

Question

For the polynomial below, -1 is a zero. $$ f(x)=x^{3}+7 x^{2}+10 x+4 $$ Express $$ f(x) $$ as a product of linear factors. $$ f(x)=\square $$

UpStudy AI · Accepted Answer

We begin with the polynomial

$$
f(x) = x^3 + 7x^2 + 10x + 4
$$

and we are given that $$-1$$ is a zero. This means that $$(x + 1)$$ is a factor of $$f(x)$$.

**Step 1: Divide $$f(x)$$ by $$(x+1)$$.**

Using synthetic division with the zero $$-1$$, we list the coefficients:

$$
1, \quad 7, \quad 10, \quad 4
$$

- Bring down the first coefficient: $$1$$.
- Multiply $$1$$ by $$-1$$: $$1 \times (-1) = -1$$.
- Add to the next coefficient: $$7 + (-1) = 6$$.
- Multiply $$6$$ by $$-1$$: $$6 \times (-1) = -6$$.
- Add to the next coefficient: $$10 + (-6) = 4$$.
- Multiply $$4$$ by $$-1$$: $$4 \times (-1) = -4$$.
- Add to the constant term: $$4 + (-4) = 0$$.

The result gives a quotient:

$$
x^2 + 6x + 4
$$

Thus, we have

$$
f(x) = (x+1)(x^2+6x+4).
$$

**Step 2: Factor the quadratic $$x^2+6x+4$$.**

To factor $$x^2 + 6x + 4$$, we use the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$

where $$a = 1$$, $$b = 6$$, and $$c = 4$$.

Compute the discriminant:

$$
\Delta = b^2 - 4ac = 6^2 - 4(1)(4) = 36 - 16 = 20.
$$

Thus,

$$
x = \frac{-6 \pm \sqrt{20}}{2} = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}.
$$

So the quadratic factors as

$$
x^2+6x+4 = \left(x-(-3+\sqrt{5})\right)\left(x-(-3-\sqrt{5})\right)= (x+3-\sqrt{5})(x+3+\sqrt{5}).
$$

**Step 3: Write the complete factorization**

Now, the polynomial can be written as the product of linear factors:

$$
f(x) = (x+1)(x+3-\sqrt{5})(x+3+\sqrt{5}).
$$
$$ f(x) = (x + 1)(x + 3 - \sqrt{5})(x + 3 + \sqrt{5}) $$

For the polynomial below, -1 is a zero.

Express as a product of linear factors.

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For the polynomial below, -1 is a zero. Express as a product of linear factors.