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## Answered question

2022-01-23

How can I convert $a{x}^{2}+bx+c=0$ to a FOIL-style $\left(x+d\right)\left(x-e\right)=0$ equation?

### Answer & Explanation

Nevaeh Jensen

Beginner2022-01-24Added 14 answers

Step 1
Theres

Rylee Marshall

Beginner2022-01-25Added 6 answers

Step 1
If a polynomial has a root r (imaginary or real), then you can pull out a factor of $x-r$.
For example, to factor ${x}^{2}+1$, we first find the roots. You can use the quadratic equation for this in general, but it's easy to see in this example that the roots are i and -i. This means the polynomial ${x}^{2}+1$ factors as $\left(x-i\right)\left(x+i\right)$.
Notice this polynomial has no real solutions. There is no amount of algebraic trickery you can do to get around that fact. Depending on your application, you will either have to accept the imaginary roots or regard the polynomial as having no (real) solutions.

RizerMix

Expert2022-01-27Added 656 answers

Step 1$a{x}^{2}+bx+c=a\left(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\right)$If you plug a number n into a polynomial and get 0, then $\left(x-n\right)$ is a factor of that polynomial. So the two numbers that you can plug into a quadratic polynomial that mkae it add up to 0 correspond to the two factors. But you can't get real numbers if the solutions are not real numbers.

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