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 ...

### 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).

### 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 ...

### How To Factor Trinomials Step By Step tutorial with practice ...

How to factor trinomials , explained with step by step examples and several practice problems.

### 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?

### 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 ...

### 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 ...

### 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 ...

### 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 ...

## Suggested Questions And Answer :

### 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

### 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.

### 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

### 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.

### 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.

### 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.

### 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