Guide :

The complete factorization of a trinomial has a product that is the trinomial.

how do you factorize a trinomial?

Research, Knowledge and Information :

Factoring trinomials - A complete course in algebra

How to factor a trinomial. Factoring ... Table of Contents | Home. 17. FACTORING TRINOMIALS. 2nd Level ... we may simply look for two numbers whose product is ...
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Factoring a Trinomial Lessons | Wyzant Resources

Factoring A Trinomial Lessons. ... This lesson explains how to factor a trinomial like this into two binomials (which we had before the FOIL Method was applied).
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Solved: Complete the factoringEXAMPLE Factoring Trinomials by ...

Answer to Complete the factoringEXAMPLE Factoring Trinomials by GroupingFactor each trinomial.(a) ... Complete the factoringEXAMPLE Factoring Trinomials by Groupi ...
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How To Factor Trinomials Step By Step tutorial with practice ...

How to factor trinomials , explained with step by step examples and several practice problems.
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Factoring Perfect Square Trinomials - MathBitsNotebook(A1 ...

Factoring Perfect Square Trinomials ... The middle term is twice the product of the binomial's ... Does this fit the pattern of a perfect square trinomial?
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Complete the factorization of 3x2 – 5x + 2. Which two factors ...

Complete the factorization of 3x2 ... Which two factors can be multiplied together to make this trinomial? (x – 2) A) (x ...
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Factoring Trinomials - Cengage

Section P.6 Factoring Trinomials 61 ... to factor a trinomial into a product of two ... the correct factorization is (b) First observe that has no common monomial ...
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The square of a binomial. Perfect square trinomials - A ...

Learn how to complete the square. ... Table of Contents | Home. 18. THE SQUARE OF A BINOMIAL Perfect square trinomials. The ... Write only the trinomial product: (x ...
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Polynomial Factoring Calculator with explanation

This online calculator writes a polynomial as a product of linear factors and creates a graph of the given ... To factor trinomial $6a^2-13ab-5b ^2$, first ...
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Suggested Questions And Answer :

The complete factorization of a trinomial has a product that is the trinomial.

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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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Factoring perfect square trinomials

Numerically, a perfect square is a number that is the product of another number multiplied with itself ( 4 x 4 = 16.  16 is a perfect square). For trinomials, a perfect square trinomial is a trinomial that is the product of a binomial multiplied with itself  (   (2x-5)(2x-5) = 4x^2 - 20x + 25.  4x^2 - 20x + 25 is a perfect square trinomial). If you know a trinomial is a perfect square going into the problem, factoring it is relatively easy.  Just find the square root of the lead term, the square root of the last term, and decide whether a + or - sign belongs in the middle. For example,  in 4x^2 - 20x + 25,  the square root of 4x^2 is 2x.  The square root of 25 is 5.  My solution will either be (2x+5)(2x+5)   or   (2x-5)(2x-5).  FOILing these out (as you should always do to check your answer) we find that the minus sign works. 4x^2 - 20x + 25 = (2x - 5)^2
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Factor the trinomial completely using the trial and error method-14y+111y+8

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what is the binomial factor 64y^15+125

64y^15+125=(4y^5)^3+5^3=(4y^5+5)(trinomial), so binomial factor=4y^5+5.
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Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer?

Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer? Our formula is x^2 + 3x = 180 Use the method called "completing the square." In order to factor the left side of the equation, we need to add the square of (1/2) * (b/a) a is the coefficient of x^2; b is the coefficient of x (1/2 * b/a)^2 = (1/2 * 3)^2 = 1.5^2 = 2.25 x^2 + 3x + 2.25 = 180 + 2.25 Factoring the left side of the equation gives (x + 1.5)*(x + 1.5) = 182.25 Notice that x * 1.5 (1.5 is 1/2 of our b/a) plus x * 1.5 gives us 3x. Also, 1.5 * 1.5 gives us 2.25 Take the square root of both sides. x + 1.5 = sqrt(182.25) The square root of 182.25 is +13.5; it is also -13.5 x + 1.5 = 13.5   and  x + 1.5 = -13.5 x = 13.5 - 1.5 = 12, not a negative number or x = -13.5 - 1.5 = -15, the number we want (-15)^2 + 3(-15) = 180 225 - 45 = 180
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relationships between sets of real numbers with visual representation

We can define some sets within the superset of all real numbers: A. Integers  A1. Positive integers or natural numbers, including zero   A1.1. Positive integers with integer exponents greater than zero.   A1.2. Negative integers with even integer exponents greater than zero  A2. Negative integers   A2.1. Negative integers with odd integer exponents greater than zero B. Fractions  B1. Proper fractions   B1.1. Positive proper fractions    B1.1.1. Positive integers with integer exponents less than zero   B1.2. Negative proper fractions    B1.2.1. Negative integers with integer exponents less than zero B2. Improper fractions (including mixed numbers)   B2.1. Positive improper fractions   B2.2. Negative improper fractions C. Irrational numbers  C1. Transcendental numbers (cannot be defined as the root of a fraction or integer)  C2. Integer root of a positive integer for integers>1  C3. Integer root of a positive fraction for integers>1  C4. Fractional root of a positive integer  C5. Fractional root of a positive fraction These are arbitrary sets that can be represented visually as circles. Some circles may be completely isolated from other circles; some may completely contain other circles; some may intersect other circles. Indentation above implies that the circle associated with the lesser indentation contains the whole of the circle with the greater indentation. For example, B1 is completely contained by B, B2.2 is contained in B2, which in turn is contained in B. The separate sets of odd and even numbers could be included. We could add the set of prime numbers, positive integers>1 with no other factors than 1 and the number itself. We could add the set of perfect numbers, integers>1 in which the factors, including 1 but excluding the number itself, add up to the number. We could add factorials (the product of consecutive integers up to the factorial integer itself). These would be included in the superset of positive integers. The visual representation is that the subset is totally enclosed by the superset as a circle inside another circle. An example of interlocking circles is the set of factorials with the set of perfect numbers, where 6 is contained in the overlap. 6 belongs to both sets.  
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how do you solve 3t^4+2t^3-300t-50=0

The expression does not factorise and the roots seem to be t=-0.16669 and 4.48653 approx. Check that you haven't missed out a term (e.g., t^2). If the equation was meant to factorise completely, then the product of the factors=-50. The factors must therefore consist of 1, 2, 5, 5. The factors of the 3t^4 must consist of 3t, t, t, t. The minus sign in front of the constant 50 implies that there must be an odd number of pluses and minuses amongst the factors: one minus and 3 pluses, or three minuses and one plus.  Let the coefficient of the apparently missing t^2 be k. The equation then reads: 3t^4+2t^3+kt^2-300t-50=0. If (t-1) is a factor then 3+2+k-300-50=0 and k=345. If (t+1) is a factor then 3-2+k+300-50=0 and k=52-303=-251. If (t-2) is a factor then 48+16+4k-600-50 and k=(1/4)(650-64)=293/2. If (t+2) is a factor then 48-16+4k+600-50 and k=(1/4)(66-648)=-291/2. It seems improbable that (t-2) or (t+2) could be a factor because k would be a mixed number (or improper fraction), so k=345 or -251. If (t-5) is a factor then 1875+250+25k-1500-50=0 and k=(1550-2125)/25=-23. If (t+5) is a factor then 1875-250+25k+1500-50=0 and k=(300-3375)/25=-123. We know that (3t+f) must be included as a factor, but if f=-5 or +5 then, since we need two 5's within the set of factors, so if one contains 3t, the other can only contain t. That means that (t-5) or (t+5) must be a factor. There's no point trying other factors like (t-25) or (t+10), because they would require (t-2) or (t+2), or (t-5) or (t+5), and (t-1) and (t+1) as other factors, which have already been covered. The exercise shows that no one value of k allows us to find more than one root. Had any two values of k above been the same we would have had at least two roots and a quadratic, which could have been solved, if it had had real roots. The conclusion is that finding one root only leaves us with a cubic equation that has no rational, real roots. This means that the question has been wrongly stated, and the inclusion of kt^2 doesn't help to solve it satisfactorily.
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how do you factor 9x^2-12x-5 completely?

You look at the factors of 9 and 5: we have 1 and 9, and 3 and 3; and we have 1 and 5. Then we look at the middle term coefficient which is 12. We note that the sign of 5 is negative, which means that we need a positive and a negative zero for the factors of the quadratic. We then try all the possibilities of combining factors of 9 and 5: so we have (1 9), (9 1) and (3 3) against (1 5) and (5 1), and we look for the combination that gives 12 when we play them off against one another. If (a b) represents the factors of 9 and (c d) the factors of 5 we're looking for the difference between ad and bc equals 12. The solution is (3 3) and (1 5) or (5 1) because 3*5-3*1=12. (Note that if the sign of 5 was positive we would be looking to find the right factors such that ad+bc=12.) Now we set up our factors in brackets: (3x 1)(3x 5) (that is, (ax c)(bx d)) without the signs. We have to end up with a minus in front of the 12 so we need to put the plus so that it's the smaller product that has the + sign, so 1 carries the plus and 5 carries the minus: (3x+1)(3x-5). This explanation sounds complicated, but it takes longer to explain it than to try it out!  Practise and you'll find it gets easier.
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the product of 2 consecutive #s is 4160 what are the numbers?

let x be the first number, then the next consecutive number is 1 more than that so it is x+1 product means multiply so x(x+1) = 4160 x^2 + x - 4160 = 0   factor this trinomial (x+ 65 )(x-64 ) = 0 set each = 0 so x + 65=0  or x = -65 then x+1 = -64 x-64=0 so x=64 then x+1 = 65 since your question does specify positive or negative, then there are two possible answers 64 and 65     or -64 and -65
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