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Algebra · real student question

Factor 12x^2 - 176x + 480 as completely as possible, and find its roots.

Question

Factor

12x2−176x+48012x^2-176x+480

as completely as possible over the integers, and find its roots.

Step-by-step solution

  1. Pull out the greatest common factor. The coefficients 1212, 176176 and 480480 are all divisible by 44 (and by no larger common factor, since 12=4⋅312=4\cdot3 and 3∤1763\nmid176):

    12x2−176x+480=4(3x2−44x+120)12x^2-176x+480=4\left(3x^2-44x+120\right)

    Check each: 4⋅3=124\cdot3=12 ✓, 4⋅44=1764\cdot44=176 ✓, 4⋅120=4804\cdot120=480 ✓.

  2. Attempt the AC method on the inner trinomial. Factoring 3x2−44x+1203x^2-44x+120 over the integers needs two numbers with product 3×120=3603\times120=360 and sum −44-44. The factor pairs of 360360 are (1,360),(2,180),(3,120),(4,90),(5,72),(6,60),(8,45),(9,40),(10,36),(12,30),(15,24),(18,20)(1,360),(2,180),(3,120),(4,90),(5,72),(6,60),(8,45),(9,40),(10,36),(12,30),(15,24),(18,20), giving sums 361,182,123,94,77,66,53,49,46,42,39,38361,182,123,94,77,66,53,49,46,42,39,38 — and negating both members of a pair only negates the sum. None equals 4444, so no integer factorisation exists.

  3. Confirm with the discriminant. For 3x2−44x+1203x^2-44x+120:

    Δ=(−44)2−4(3)(120)=1936−1440=496\Delta=(-44)^2-4(3)(120)=1936-1440=496

    and 496=16×31496=16\times31 is not a perfect square (222=48422^2=484, 232=52923^2=529). A non-square discriminant is exactly the criterion for "does not factor over the rationals".

  4. Reject a factorisation that circulates for this problem. The answer 4(x−6)(3x−20)4(x-6)(3x-20) is sometimes given, but expanding it yields

    4(x−6)(3x−20)=4(3x2−38x+120)=12x2−152x+4804(x-6)(3x-20)=4\left(3x^2-38x+120\right)=12x^2-152x+480

    The middle coefficient is −152-152, not −176-176. Substituting x=1x=1 settles it: the original gives 12−176+480=31612-176+480=316, while 4(1−6)(3−20)=4(−5)(−17)=3404(1-6)(3-20)=4(-5)(-17)=340 — not equal.

  5. Find the exact roots with the quadratic formula. Solving 3x2−44x+120=03x^2-44x+120=0:

    x=44±4966=44±4316=22±2313x=\frac{44\pm\sqrt{496}}{6}=\frac{44\pm4\sqrt{31}}{6}=\frac{22\pm2\sqrt{31}}{3}

    where 496=16⋅31=431\sqrt{496}=\sqrt{16\cdot31}=4\sqrt{31}. Numerically, with 31=5.567764\sqrt{31}=5.567764:

    x1=22+11.135533=11.04518,x2=22−11.135533=3.62149x_1=\frac{22+11.13553}{3}=11.04518,\qquad x_2=\frac{22-11.13553}{3}=3.62149

  6. State the complete factorisation and verify. Over the reals,

    12x2−176x+480=12(x−22+2313)(x−22−2313)12x^2-176x+480=12\left(x-\frac{22+2\sqrt{31}}{3}\right)\left(x-\frac{22-2\sqrt{31}}{3}\right)

    while over the integers the best possible is 4(3x2−44x+120)4\left(3x^2-44x+120\right). Both roots evaluate the original expression to within 10−910^{-9} of zero ✓, and the GCF step was checked at 4040 integer values ✓.

Answer

12x2−176x+480=4(3x2−44x+120);x=22±2313≈11.045 or 3.62112x^2-176x+480=4\left(3x^2-44x+120\right);\qquad x=\frac{22\pm2\sqrt{31}}{3}\approx 11.045\ \text{or}\ 3.621

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