Can all quadratic equations be factorised?

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SUMMARY

Not all quadratic equations of the form ax² + bx + c can be factorized into the product (px + q)(rx + s) without specific conditions. If q and s are allowed to be complex numbers, factorization is always possible. However, if q and s are restricted to real numbers, the condition √(b² - 4ac) ≥ 0 must be satisfied. For rational numbers, √(b² - 4ac) must be a perfect square for factorization to occur.

PREREQUISITES
  • Understanding of quadratic equations and their standard form (ax² + bx + c)
  • Knowledge of complex and real numbers
  • Familiarity with the discriminant (b² - 4ac) and its implications
  • Basic algebraic factorization techniques
NEXT STEPS
  • Research the properties of complex numbers in algebra
  • Study the discriminant of quadratic equations and its role in factorization
  • Learn about perfect squares and their significance in rational factorization
  • Explore advanced factorization techniques for polynomials
USEFUL FOR

Students studying algebra, educators teaching quadratic equations, and mathematicians interested in polynomial factorization methods.

Kyoma
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Homework Statement



Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.
 
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Kyoma said:

Homework Statement



Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.

If d is just some arbitrary number and you didn't accidentally duplicate your letters, then no.

Example:

y = x2-2x+1

So a = 1, b = -2, c = 1

So y = (x-1)(x-1)

a, c and d are fine, but you have a -1 instead of b.
 
Kyoma said:

Homework Statement



Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.
As gb7nash said the coefficients a, b, and c, in (ax2+bx+c) are not, in general, the same as those in (ax+b)(cx+d).

So, can (ax2+bx+c) always be factored into the form, (px+q)(rx+s)?
The answer is ... maybe.

If you allow q and s to be complex numbers, then the answer is yes.

If you restrict q and s to be real numbers, then you must have [itex]\sqrt{b^2-4ac}\ge0[/itex]

If q and s must be rational numbers, then you must have [itex]\sqrt{b^2-4ac}[/itex] be a perfect square.​

This all assumes that a, b, and c are rational numbers.
 

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