请求对二次多项式2x²-5x+3因式分解及分式化简指导

Got it, let's walk through this step by step so you can fully simplify that quadratic fraction. We'll start with the denominator you're stuck on, then handle the numerator, and finally cancel out common factors.

Step 1: Factorizing the Denominator (2x^2 -5x +3) #

For quadratics where the leading coefficient isn't 1 (here, (a=2)), the cross-multiplication method works really well:

• First, multiply the leading coefficient ((2)) and the constant term ((3)): (2*3=6)
• Find two numbers that multiply to (6) and add up to the middle coefficient ((-5)). Those numbers are (-2) and (-3) (since (-2*-3=6) and (-2 + (-3)=-5))
• Rewrite the middle term using these two numbers: (2x^2 -2x -3x +3)
• Group the terms and factor out the GCF from each pair:
  • (2x(x -1) -3(x -1))
• Now factor out the common binomial ((x-1)):
  • ((2x -3)(x -1))

You can double-check by expanding: ((2x-3)(x-1) = 2x^2 -2x -3x +3 = 2x^2 -5x +3) — that's exactly our original denominator.

Step 2: Factorizing the Numerator (x^2 +3x -4) #

This is a standard monic quadratic (leading coefficient is 1):

• Find two numbers that multiply to (-4) and add up to (3). That's (4) and (-1) (since (4*-1=-4) and (4 + (-1)=3))
• Factor it directly as:
  • ((x +4)(x -1))

Verify by expanding: ((x+4)(x-1) = x^2 -x +4x -4 = x^2 +3x -4) — perfect, that matches the numerator.

Step 3: Full Simplification of the Rational Expression #

Substitute both factorizations back into the original fraction:

(x² +3x -4)/(2x² -5x +3) = [(x+4)(x-1)] / [(2x-3)(x-1)]

The critical part: we can cancel the common factor ((x-1)) only if (x≠1) (since (x=1) would make the original denominator zero, which is undefined). After canceling, we get:

(x +4)/(2x -3)

Don’t forget the restrictions on (x): (x≠1) and (x≠3/2) (since (2x-3=0) when (x=3/2), which also makes the denominator zero).

内容的提问来源于stack exchange，提问作者Anonymous Anonymous