what is the completed factored form of this expression?    4b (5-a) -3c (5-a)?
448387
4b(5 - a) - 3c(5 - a)\n= (5 - a)(4b - 3c)\n\nThe answer is D. Since both have the factor (5 - a), you factor it out, and what's left is (4b - 3c).
737767
No, it is a real number. \n\nTo factor this write it as \n\n(x - 3)^2 - 5\n\nThis is the difference of 2 squares, (x - 3) and sqrt(5). So the factorization is \n\n((x - 3) + sqrt(5))*((x - 3) - sqrt(5))\n\nIn this form it is easy to see if the answer is real or not, if the right side is nonnegative there are real solutions, if it is < 0 then there are none.
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
677061
i think this formula works.\nfactorial of A-4A\nso.using this formula the correct answer is 5012
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
600095
You can factor out 15bc^2\n15bc^2(5b + 4c^4)
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
767628
I'm assuming you want to factor this\n\n5z^2 + 4z -12 = (5z -6)(z+2)
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
136651
7c\n\n\n2*7*c*c\n\n5*7*c\n\ncommon factors:  7 and c\n\ntherefore, it's 7c
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
1070097
(x+3) (5x-3) (7x-1)\n\nYou can do this manually by figuring out the factors of 35 and 9.\n\nFor 35:  {1, 5, 7, 35, -1, -5, -7, -35)\nFor 9: {1, 3, 9, -1, -3, -9)\n\nI started assuming that (x + n) would be a factor.  I tried (x + 1), (x - 1), then (x + 3).\n\nIf you do long division, you get:\n(x + 3) ( 35x^2 - 26x^2 + 3)\n\nThen you just repeat the process for ( 35x^2 - 26x^2 + 3)... you need factors of 35 and factors of 3 that will add up to -26.  \n-3 x 7 = -21 and -1 x 5 = -5, so (5x - 3)(7x - 1) works.\n\nThat gives the final answer of: \n(x+3) (5x-3) (7x-1)
6182
You can use what is called the Master Product Method to factor this and other quadratic trinomials.\n\nFor something of the form ax^2 + bx + c, the master product is a*c.  In this example, a is 2, b is -5 and c is 2, so the master product is 2*2 = 4.  Now you must find two numbers which multiply to give you the master product of 4, but add to give the middle coefficient of -5.\n\nThis requires a bit of trial and error, but you should be able to quickly discover that the numbers -1 and -4 will work since -1*-4 = 4 and -1+-4 = -5.  Now you rewrite the the entire trinomial replacing the middle term -5x with the equivalent -1x - 4x:\n\n    2x^2 - 5x + 2\n= 2x^2 - 1x - 4x + 2\n\nWe factor by grouping the first pair and last pair together and pulling out what's common to each pair:\n\n= (2x^2 - 1x) + (-4x + 2)\n= x(2x - 1) -2(2x - 1)\n\nNotice that (2x-1) appears in the factorization of both pairs.  We finish by factoring out the (2x-1) to get\n\n= (x - 2)(2x -1)
866906
q = (5r-3) / (4r-1) multiply by 4r-1\n4qr-q=5r-3 add q\n4qr=5r-3+q    subtract 5r\n4qr-5r=q-3 factor out r\nr(4q-5)=q-3   divide by 4q-5\nr=(q-3)/(4q-5)
534750
1/2 + 1/2b + 2b^2 + 6bc + c^2\n\n(1/4b)(1/4b) + 4bc + (bc)(bc)\n1/2 + 1/4b +1/4b + b^2 +4bc + b^2 + c^2 + 2bc\n1/2 + 1/2 + 2b^2 + 6bc + c^2\n\nI think this is right... email me if I'm wrong?
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
504201
Actually, none of the above are correct. To factor anything in the form of a cubed - b cubed, where a and b are real numbers, the formula is (a - b)(a^2 + ab + b^2). In this case, the a would be 1 and the b would be 4. So, the answer would be (x-4)(x^2 + 4x + 16i)\n\nand paul is incorrect also, try to multiple it out, it's wrong