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Algebra · real student question

Factor x^2 - 3x + 2, then use the factorisation to solve x^2 - 3x + 2 = 0.

Question

Factor

x2−3x+2x^2-3x+2

and then use the factorisation to solve x2−3x+2=0x^2-3x+2=0.

Step-by-step solution

  1. Read the signs off the coefficients first. The constant +2+2 is positive, so the two factors share a sign; the middle coefficient −3-3 is negative, so that shared sign is negative. We are looking for two negative numbers.

  2. Find the pair. With a leading coefficient of 11, the numbers must multiply to 22 and add to −3-3:

    (−1)⋅(−2)=2,−1+(−2)=−3(-1)\cdot(-2)=2,\qquad -1+(-2)=-3

    so the pair is −1-1 and −2-2.

  3. Write the factorisation.

    x2−3x+2=(x−1)(x−2)x^2-3x+2=(x-1)(x-2)

  4. Solve the equation with the zero-product property. A product is zero exactly when one factor is zero:

    x−1=0 ⇒ x=1,x−2=0 ⇒ x=2x-1=0\ \Rightarrow\ x=1,\qquad x-2=0\ \Rightarrow\ x=2

    Note the sign flip: the factor (x−1)(x-1) gives the positive root 11.

  5. Check both the factorisation and the roots. Expanding: (x−1)(x−2)=x2−2x−x+2=x2−3x+2(x-1)(x-2)=x^2-2x-x+2=x^2-3x+2 ✓\checkmark. Substituting: at x=1x=1, 1−3+2=01-3+2=0 ✓\checkmark; at x=2x=2, 4−6+2=04-6+2=0 ✓\checkmark. Vieta agrees too, since the roots sum to 3=−ba3=-\tfrac{b}{a} and multiply to 2=ca2=\tfrac{c}{a}.

Answer

x2−3x+2=(x−1)(x−2),x=1 or x=2x^2-3x+2=(x-1)(x-2),\qquad x=1\ \text{or}\ x=2

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