Guide :

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

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

## Research, Knowledge and Information :

### 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).

### 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 ...

### 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 ...

### 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 ...

### 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

### 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 ...

### Factor 2x 2 - 10x + 8 completely - Brainly.com

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 ...

### 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 ...

## 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).

### 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).

### 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))

### 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).

### 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).

### Factorise 18uv-9v

QUESTION: Factorise 18uv-9v . 18uv-9v factorises as: 9v(2u - 1)

### 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)

### 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.)

### 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)