# What is k(x)= 2x^3-5x^2-x+6 written in standard form and as a product of linear factors?

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We have the equation

`k(x) = 2x^3 - 5x^2 -x + 6`

To write this in standard form, set the equation to 0. Then we can solve for the roots by writing the equation as a product of linear factors.

To write the equation as a product of linear factors, that is factors that contain terms in x with an power/exponent of at most 1, of the form `x+a` .

This equation is a cubic since the largest power in x is 3. For quadratics, where the largest power in x is 2, we usually use the quadratic formula to express the equation as a product of two linear factors. However, for cubics the formula is much more complex and not used for hand calculations, only by computers.

Instead, we look at the constant term, as we might if factorising a quadratic by inspection: the constant term will be the product of the constants in the n linear factors, where n is the order of the polynomial (for a cubic, which we have here, n =3). Our constant term here is 6. To guess the constants in the 3 linear terms we need to find, we need to consider the possible factors of 6. The constants could be non-integer, but with hand calculations set as questions, the constants would tend to be 'nice' integers, so we would begin with those.

So, 6 has integer factors `pm` 1,`pm` 2, `pm` 3 and `pm` 6. The roots of our cubic equation would most likely be these, though could also be non-integer numbers. There could be repeated roots. We need to work out the roots by exhaustion, trying each in turn, in the absence of a rigorous formula to plug the coefficients of the terms in x into.

Try x=1: 2 - 5 - 1 + 6 `!= 0` . Not a root

x=2: 16 - 20 - 2 + 6 = 0. A root!

As soon as we find a root, we can factor the associated linear factor out, by writing the equation as a product of '(x-root)' and a quadratic expression in x:

`2x^3 - 5x^2 - x + 6 = (x-2)(2x^2-x-3)` = 0

To find the other 2 roots, we can now solve the quadratic expression obtained, either by using the quadratic formula by inspection. This is slightly trickier than the most basic quadratics to solve by inspection, as the coefficient of the `x^2` term is 2 and not the standard value of 1. Thinking again, as with the original cubic, about factors of the constant term, but now also about the coefficient of the term in x, we need to consider integers `a` and `b` that solve the pair of equations

`-(2a+b) = -1````` and `ab = -3`  simultaneously, where the first equation ensures the coefficient in x is -1 and the second ensures that the constant term is -3.The remaining two linear factors of the original cubic would then be of the form (2x-b) and (x-a) respectively. Looking at the second of these equations, `a` and `b` could only be 1,-1,3,-3 as these are the only integer solutions. Without going through these exhaustively (if possible, as this speeds things up), it can be seen that `` `a=-1` and ```b=3` is the solution we're after. The pair `(a,b/2)` `` are the other two roots of the original cubic equation. Check these by a quick hand calculation:

x = -1: -2 - 5 + 1 + 6 = 0. A root.

x = 3/2 : 2(27/8) - 5(9/4) - (3/2) + 6 = 27/4 - 45/4 - 6/4 + 24/4 = 0. A root.

Finally, we can write the original cubic as a product of three linear factors:

`k(x) = 0`  `implies` `(x-2)(2x-3)(x+1) = 0` ````

The three roots of the equation can be found by setting each of the linear factors to zero, implying the roots are x = 2, 3/2, -1. These are the points on the x axis where the cubic graph crosses/cuts the axis.