For the polynomial below, 1 is a zero of multiplicity two.

Express as a product of linear factors.

Ask by Hill Luna. in the United States

Mar 29,2025

Question

For the polynomial below, 1 is a zero of multiplicity two. $$ h(x)=x^{4}-4 x^{3}+10 x^{2}-12 x+5 $$ Express $$ h(x) $$ as a product of linear factors. $$ h(x)=\square $$

UpStudy AI · Accepted Answer

$$
h(x)= (x-1)^2\,(x-(1+2i))\,(x-(1-2i))
$$

To explain the steps:

1. Since $$ x=1 $$ is a zero of multiplicity two, we factor out $$(x-1)^2$$.

2. We write
$$
h(x)=(x-1)^2 Q(x)
$$
where $$ Q(x) $$ is quadratic. Let
$$
Q(x)=ax^2+bx+c.
$$

3. Expanding, we have
$$
(x-1)^2(ax^2+bx+c)= (x^2-2x+1)(ax^2+bx+c)= ax^4+(b-2a)x^3+(c-2b+a)x^2+(b-2c)x+c.
$$

4. Equate the coefficients with
$$
x^4-4x^3+10x^2-12x+5:
$$
   - For $$ x^4 $$: $$ a=1 $$.
   - For $$ x^3 $$: $$ b-2= -4 $$ hence $$ b=-2 $$.
   - For $$ x^2 $$: $$ c-2(-2)+1 = c+4+1 = c+5=10 $$ hence $$ c=5 $$.

5. Thus, 
$$
Q(x)= x^2-2x+5.
$$

6. The quadratic $$ Q(x)=x^2-2x+5 $$ has discriminant
$$
D=(-2)^2-4(1)(5)=4-20=-16,
$$
so its roots are complex:
$$
x= \frac{2\pm \sqrt{-16}}{2} = 1\pm 2i.
$$

7. Therefore, the complete factorization into linear factors is:
$$
h(x)=(x-1)^2\,(x-(1+2i))\,(x-(1-2i)).
$$
$$ h(x) = (x - 1)^2 (x - (1 + 2i))(x - (1 - 2i)) $$

For the polynomial below, 1 is a zero of multiplicity two.

Express as a product of linear factors.

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