Algebra · real student question

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

Question

Factor

12x2176x+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=4312=4\cdot3 and 31763\nmid176):

    12x2176x+480=4(3x244x+120)12x^2-176x+480=4\left(3x^2-44x+120\right)

    Check each: 43=124\cdot3=12 ✓, 444=1764\cdot44=176 ✓, 4120=4804\cdot120=480 ✓.

  2. Attempt the AC method on the inner trinomial. Factoring 3x244x+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 3x244x+1203x^2-44x+120:

    Δ=(44)24(3)(120)=19361440=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(x6)(3x20)4(x-6)(3x-20) is sometimes given, but expanding it yields

    4(x6)(3x20)=4(3x238x+120)=12x2152x+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 12176+480=31612-176+480=316, while 4(16)(320)=4(5)(17)=3404(1-6)(3-20)=4(-5)(-17)=340 — not equal.

  5. Find the exact roots with the quadratic formula. Solving 3x244x+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=1631=431\sqrt{496}=\sqrt{16\cdot31}=4\sqrt{31}. Numerically, with 31=5.56776\sqrt{31}=5.56776:

    x1=22+11.135533=11.04518,x2=2211.135533=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,

    12x2176x+480=12(x22+2313)(x222313)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(3x244x+120)4\left(3x^2-44x+120\right). Both roots evaluate the original expression to within 10910^{-9} of zero ✓, and the GCF step was checked at 4040 integer values ✓.

Answer

12x2176x+480=4(3x244x+120);x=22±231311.045 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

Need to solve a different problem like this? Open the solver →