# Why can $(\lambda - a)(\lambda - d)-bc = 0$ be rewritten as $(\lambda - \lambda_1)(\lambda - \lambda_2) = 0$?

I have $$(\lambda - a)(\lambda - d)-bc = 0$$ which can also be written as $$\lambda^2 - \lambda(a+d) + (ad - bc) = 0$$. This quadratic equation can be solved by finding the roots $$\lambda_1$$ and $$\lambda_2$$. With the roots, I have been told that $$\lambda^2 - \lambda(a+d) + (ad - bc) = 0$$ can be written as $$(\lambda - \lambda_1)(\lambda - \lambda_2) = 0$$ - why?

• You said $(\lambda - a)(\lambda - d)-bc = 0$ can be written as $\lambda^2 - \lambda(a+b) + (ad - bc) = 0$ but shouldn't it be $-\lambda(a+d)$? – ArsenBerk Jun 6 '19 at 11:36
• Yes, of course. Thanks! – Fac Pam Jun 6 '19 at 11:43
• Is your question related to chromatic numbers? Because you are used $\lambda$ as variable. – BarzanHayati Jun 6 '19 at 12:48
• It's obvious, you could write an equation with two roots in two forms. So your question is not clear. – BarzanHayati Jun 6 '19 at 12:50

If $$x=\alpha$$ is a root of a polynomial $$p(x)$$, you can always write $$p(x)$$ as $$p(x) = (x-\alpha)\cdot q(x),$$ where $$q(x)$$ is a polynomial with degree $$\deg(q)=\deg(p)-1$$.

In your example $$p(\lambda) = (\lambda - a)(\lambda - d)-bc$$ is a polynomial of degree $$2$$. Since it has roots $$\lambda_1$$, $$\lambda_2$$ you can write is as $$p(\lambda) = (\lambda-\lambda_1)(\lambda-\lambda_2)\cdot q(\lambda)$$ where $$q(\lambda)$$ has degree $$0$$, that is, is a constant. However, since the coefficient of $$\lambda^2$$ in $$p(\lambda)$$ is $$1$$ we must have $$q(\lambda)=1$$ so that $$p(\lambda) = (\lambda-\lambda_1)(\lambda-\lambda_2).$$

Hint: Expanding $$(\lambda-a)(\lambda-d)-bc=0$$ we get $$\lambda^2-a\lambda-d\lambda+ad-bc=0$$ this can be written as $$\lambda-\lambda(a+d)+ad-bc=0$$ using the quadratic formula we obtain $$\lambda_{1,2}=\frac{a+d}{2}\pm\frac{1}{2}\sqrt{(a-d)^2+4bc}$$ now you can write $$(\lambda-\lambda_1)(\lambda-\lambda_2)=0$$ It is the same like above, called the theorem of Vieta.

• What does this have to do with the roots $\lambda_1$ and $\lambda_2$? – Fac Pam Jun 6 '19 at 11:31
• I still do not understand why you can get from $$\lambda_{1,2}=\frac{a+d}{2}\pm\frac{1}{2}\sqrt{(a-d)^2+4bc}$$ to $$(\lambda-\lambda_1)(\lambda-\lambda_2)=0$$ – Fac Pam Jun 6 '19 at 11:39
• @FacPam See the BiFactor Theorem. and the higher degree generalization of the Factor Theorem that I prove there. – Bill Dubuque Jun 6 '19 at 17:04

Referring to Vieta's theorem: $$\lambda^2 - \lambda(a+b) + (ad - bc) = 0 \Rightarrow \begin{cases}\lambda_1+\lambda_2=a+b\\ \lambda_1\lambda_2=ad-bc\end{cases}$$ and: $$(\lambda - \lambda_1)(\lambda - \lambda_2) = 0 \Rightarrow \lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2=0$$ Do you see the relationship?

Have a close look at the polynomial $$p_α(λ) = α(\lambda - λ_1)(\lambda - λ_2)$$, $$α\ne0$$. Its roots are really obvious: $$p_α(λ) = 0$$ if and only if at least one of the factors $$λ-λ_1$$ and $$λ-λ_2$$ is zero. So this polynomials have the same roots as your polynomial $$p(λ) = (\lambda - a)(\lambda - d)-bc$$. The correct value of $$α$$ is the leading coefficient of $$p(λ) = 1\cdotλ^2−λ(a+d)+(ad−bc)$$, that is $$α=1$$.

When you really want to understand the mathematical background why this factorization works you need the Fundamental Theorem of Algebra but this may be beyond your skills.