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.

Question

UpStudy AI · Accepted Answer

We know that $$1$$ is a zero of multiplicity two for the polynomial $$h(x) = x^{4} - 4x^{3} + 10x^{2} - 12x + 5$$. This means that $$(x - 1)^{2}$$ is a factor of $$h(x)$$.
To express $$h(x)$$ as a product of linear factors, we will first divide $$h(x)$$ by $$(x - 1)^{2}$$ to find the other factors.
Factor the expression by following steps:
- step0: Factor:
  $$x^{4}-4x^{3}+10x^{2}-12x+5$$
- step1: Calculate:
  $$x^{4}-2x^{3}+5x^{2}-2x^{3}+4x^{2}-10x+x^{2}-2x+5$$
- step2: Rewrite the expression:
  $$x^{2}\times x^{2}-x^{2}\times 2x+x^{2}\times 5-2x\times x^{2}+2x\times 2x-2x\times 5+x^{2}-2x+5$$
- step3: Factor the expression:
  $$x^{2}\left(x^{2}-2x+5\right)-2x\left(x^{2}-2x+5\right)+x^{2}-2x+5$$
- step4: Factor the expression:
  $$\left(x^{2}-2x+1\right)\left(x^{2}-2x+5\right)$$
- step5: Factor the expression:
  $$\left(x-1\right)^{2}\left(x^{2}-2x+5\right)$$
After performing the polynomial division, we find that $$h(x)$$ can be expressed as $$h(x) = (x - 1)^{2}(x^{2} - 2x + 5)$$.
Next, we need to factor the quadratic expression $$x^{2} - 2x + 5$$. We will check if it can be factored further.
The quadratic expression $$x^{2} - 2x + 5$$ does not factor into real linear factors, as its discriminant is negative. Therefore, it remains as is.
The polynomial $$h(x)$$ can be expressed as a product of linear factors as follows: $$ h(x) = (x - 1)^{2}(x^{2} - 2x + 5) $$ The quadratic part $$x^{2} - 2x + 5$$ does not factor further into real linear factors.

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

Express as a product of linear factors.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Algebra Questions

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