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I suppose you mean factoring the polynomial. You can check by multiplying the factors - the result should be the original polynomial.
The general form of a quadratic expression is given as ax2+bx+c where "a" cannot equal zero and "b" is the coefficient of the variable "x" and also the sum of the factors of "c" when "a" is unity. Example: x2+5x+6 = (x+2)(x+3) when factored
It doesn't look as if you can solve this easily with factoring; you might try completing the square, or use the quadratic formula, with a = 1, b = 34, c = 56.Improved answer:Presumably this is a quadratic equation in the form of x2+34x = 56.Rearrange the equation in the form of:x2+34x-56 = 0Then by completing the square or using the quadratic equation formula the values of x will work out as:x = -17- the square root of 345or x = -17+ the square root of 345Your maths tutor should be familiar with the above methods of solving quadratic equations if you're not too sure.
You should look for two numbers whose product is 40, and whose sum is 13. Experimenting a little (trying different pairs of factors for 40), you can quickly realize that this works for 8 and 5. Therefore, x2 + 13x + 40 = (x + 8) (x + 5) that would be 8 and 5 If you want to use the quadratic equation to solve it instead of factoring, the equation is x = (-b + or - the square root of (b2- 4ac))/2a a in this case is 1, b is 13, and c is 40 So, plug a, b, and c into the above equation and do it once with adding and once with subtracting. There will be two x values given if you do it both times, 8 and 5.
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I suppose you mean factoring the polynomial. You can check by multiplying the factors - the result should be the original polynomial.
-- If the last term of the trinomial ... the one that's just a number with no 'x' ... is positive, then both factors have the same sign as the middle term of the trinomial. -- If the last term is negative, then the factors have different signs. If this was never pointed out in class, well, it should have been.
waht are the special features a factoring company should have
The general form of a quadratic expression is given as ax2+bx+c where "a" cannot equal zero and "b" is the coefficient of the variable "x" and also the sum of the factors of "c" when "a" is unity. Example: x2+5x+6 = (x+2)(x+3) when factored
Is the coefficient of the square a prime number? eg if the equation begins 3a2 then the factors must be (3a +/- x)(a +/- y)
In doing the factor tree, you can start with any two factors of the given number. These factors maybe prime or composite. But the resulting factors in the factor tree should always be prime numbers. It means that you have to continue factoring unless you ended up with all prime factors.
Factoring services can be very useful tools for truck drivers to use. They should look into services, such as bill or invoice factoring, because they may end up needing them.
For the purpose of this discussion, you should consider the terms "x-intercepts," "roots," "solutions" and "zeroes" to be interchangeable. Let's take a couple of random x- intercepts, -3 and 5. For those to have been the solutions, they had to have come from (x + 3) and (x - 5) respectively. If you multiply them together, you'll get the quadratic equation x2 - 2x - 15Essentially, what we've done here is the factoring process backwards.
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It doesn't look as if you can solve this easily with factoring; you might try completing the square, or use the quadratic formula, with a = 1, b = 34, c = 56.Improved answer:Presumably this is a quadratic equation in the form of x2+34x = 56.Rearrange the equation in the form of:x2+34x-56 = 0Then by completing the square or using the quadratic equation formula the values of x will work out as:x = -17- the square root of 345or x = -17+ the square root of 345Your maths tutor should be familiar with the above methods of solving quadratic equations if you're not too sure.
The function on a ti-89 that gives you the zeros of a quadratic equation is called just that "zeros". To access it from the home screen, press f2 and select the label called "zeros(" then type the function and define the variable. For example: if you want the zeros of y=x^2+7x+12 you the display should read: zeros(x^2+7x+12,x), press enter and it will give you the results in this case {-3, -4}. We can check if it did it right by factoring this simple quadratic. 0=x^2+7x+12 factors as 0=(x+3)(x+4) set the factors equal to zero: x+3=0 x=-3 x+4=0 x=-4 So we see that the calculator did it right! That is always a good thing. This will work for most polynomial functions.