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how to factorise 2x^2+10x-8

i have to factorise this quadratic equation- 2x^2+10x-8

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Equation Factoring Calculator | Wyzant Resources

Equation Factoring Calculator. Equation: Variable: ... Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5).
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How do you factor completely: 10x^2 + 2x − 8? | Socratic

How do you factor completely: #10x^2 + 2x − 8#? Algebra Polynomials and Factoring Factoring Completely. 1 Answer ? 2 Nghi N ...
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3x^2+10x+8=0 - Get Easy Solution

Simplifying 3x 2 + 10x + 8 = 0 Reorder the terms: 8 + 10x + 3x 2 = 0 Solving 8 + 10x + 3x 2 = 0 Solving for variable 'x'. Factor a trinomial. ... Set the factor '(2 ...
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x^2-10x-8=0 - solution - Get Easy Solution

Simple and best practice solution for x^2-10x-8=0 equation. Check how easy it is, ... 2x-2=8 x-3=5 3x+2=18 2x+10=12 6x-2=14 3x=12 4x-2=12 9x-3=6 12+x=5 x+8=13 all ...
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factor 3x^2+10x-8 - OpenStudy

factor 3x^2+10x-8(3x+4)(x+2). (3x-2)(x+4). sorry, i read the -8 wrong. Nancy is right! anonymous 6 years ago; factor ... factor 3x^2+10x-8. Mathematics
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Factorise 3x^2 - 10x - 8? - Weknowtheanswer

Factorise 3x^2 - 10x - 8? Find answers now! ... So you keep fiddling around, and eventually you find that (3x+2)(x-4) = 3x^2 - 12x + 2x - 8 = 3x^2 - 10x - 8. By the ...
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Factor 2x 2 - 10x + 8 completely -

Factor 2x 2 - 10x + 8 completely 1. Ask ... 16 and we also know that 2 + 8 = 10 so this is perfect factor for this equation. 2x² - 2x - 8x + 8 Step 2 Talking commons ...
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How do you factor 2x^2 - 10x - 12? | Socratic

(2x+2)(x -6) is the factorised form of the expression. 2x^2 -10x-12 We can Split the Middle Term of this expression to factorise it. ... How do you factor #-x^2 + x ...
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Suggested Questions And Answer :

factorise 6x^2+14x-8

2(3x^2+7x-4) doesn't factorise rationally any further but 2(3x^2-7x+4) does factorise: 2(3x-4)(x-1); also 2(3x^2+7x+4) factorises: 2(3x+4)(x+1).  
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factorise x^3+(1/x^3)-52

x^3+(1/x^3) as part of the expression factorises into (x+(1/x))((x^2)-1+(1/x^2)), but 52 factorises into {26*2} or {13*4} only. The whole expression can be written (x^6+1-52x^3)/(x^3). If we let y=x^3, this becomes (y^2-52y+1)/y. The quadratic expression for y in the numerator doesn't factorise into rational terms. It can be written (y-26+15sqrt(3))(y-26-15sqrt(3))/y. Then we can put x^3 back in place of y. This makes the factors x-2-sqrt(3) and x-2+sqrt(3).
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how do you combine like terms with the same variable?

Like terms are where the exponents and the variable are the same but the coefficients may be different. Constants are all considered to be like terms. The same variable with different exponents can only be combined by factorisation, the variable with the lowest exponent usually being a common factor. Factorisation can also be used when two coefficients have a common factor. You can't combine different variables, even if they have the same exponent. Factorisation can be used when you have products of different variables. EXAMPLE 16+3x^3+5x^2+2x^3-y^2-x+4y+8x+y+6y^2-2x+9-25 can be simplified: x^3 term: 3x^3+2x^3 is 5x^3 x^2 term: 5x^2 only y^2 term: -y^2+6y^2 is 5y^2 x term: -x+8x-2x is 5x y term: 4y+y is 5y constant: 16+9-25 is 0 So we have: 5x^3+5x^2+5y^2+5x+5y. But 5 is a common factor: 5(x^3+x^2+y^2+x+y). Further factorisation is possible but looks clumsy: 5(x(x^2+x+1)+y(y+1))      
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factorise the sum x^3-x^2+x+1

This cubic expression doesn't factorise although it has a zero at about -0.5437. The question arises: are the + and - correct? For example: x^3-x^2-x+1 does factorise: (x+1)(x-1)^2 and x^3-x^2+x-1 factorises: (x-1)(x^2+1) or further (x-1)(x+i)(x-i).
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how to factorise m2+ 3mn-4n2 completely and withot like terms left

The expression is really asking what two numbers when multiplied come to 4 and when subtracted from one another come to 3? The answer is, of course, 4 and 1. So that starts the factorisation. The answer must look like (m+an)(m-bn). When we expand this we get: m^2-bmn+amn-abn^2. This can be written m^2+(a-b)mn-abn^2. We already know what a and b are: they're 1 and 4, but which way round? Looking at the expression again we see the middle term has a coefficient of +3, so a-b=3; therefore a must be bigger than b, making a=4 and b=1. This makes the factorisation (m+4n)(m-n).
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Factorise 18uv-9v

QUESTION: Factorise 18uv-9v . 18uv-9v factorises as: 9v(2u - 1)
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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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how can we factorise 1) 4x^2+3x-1 and 2) 2x^3-250

We need factors of 4 whose difference is 3: 4 and 1 fit. So factorise: (4x-1)(x+1). Take out the common factor 2: 2(x^3-125), then we can see that x=5 is a zero of the cubic, so (x-5) is a factor and x^3-125 can be divided by x-5 to find the other factor: x^2+5x+25. The solution is 2(x-5)(x^2+5x+25)
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What are the characteristics of a quadratic equation with two solutions,one solution,and no solution?

Descibe the quadratic equation with two solutions,one solution and no solution. All quadratic equations have two solutions. When there is only one solution, there are also two solutions. Such a solution is called a double root. When there is no solution, there are also two solutions, both of them complex. Any quadratic equation can be factorised. Let the quadratic be ax^2 + bx + c = 0 Sometines, this expression can be simply factorised into (x - r1)(x - r2) = 0. In such a case, you will have two real roots, x = r1, and x = r2. Sometimes the expression can be factorised into a complete square, e.g. (x - r)^2 = 0. This is when you will have a double root, x = r (twice). For the complex case, this will mean that the discriminant is negative. i.e. b^2 - 4ac < 0. You will now still have two solutions, They will be of the form: x = u + i.v, and x = u - i.v, where i = sqrt(-1). N.B. note that the above two solutions are complex conjugates . (They will always be so.)
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factorise 2mk - m - p (1 - 2k) by grouping

Question: factorise 2mk - m - p (1 - 2k) by grouping. 2mk - m - p (1 - 2k)    ---  take out a factor of m from the 1st two terms m(2k - 1) - p(1 - 2k)   ---  change the sign of the 1st product -m(1 - 2k) - p(1 - 2k)  ---  take out the common factor of (1 - 2k) (1 - 2k)(-m - p)           ---  adjust the sign (2k - 1)(m + p)
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