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# How to solve this maths problem, X^4-3X^3+4X^2-8

I need factors for that maths problem.Thank you very much for the support

Nish, make sure you copy the problem exactly. Thanks

The problem is this

X^4-3X^3+4X^2-8

this is a past paper question

### 2 Answers by Expert Tutors

Marc S. | Polymath tutoring math, science, and writingPolymath tutoring math, science, and wri...
4.9 4.9 (697 lesson ratings) (697)
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Hi Nish, and welcome to the wonderful world of factoring polynomials of degree 3 and above. That wild beast is a quartic because it's degree 4, and here's how I would try to tame it by factoring it into nice domesticated polynomials of degree 1 and 2. These are the the linear and quadratic polynomials that you've worked with in the past.

The first step is to see if all four terms have a factor in common. They don't because that would just be way too easy. In algebra 1, all four terms would have had something in common.

The second step is to see if factoring by grouping will work, so take a look at (x4−x3)+(4x2−8). What factors do the first two terms have in common? An x3 term. And what factors do the second two terms have in common? Just a little 4. Factor those expressions one at a time, and you're left with:

x3(x−1)+4(x2−2).

Uh oh. In algebra 1 those expressions in parentheses would have been identical, and now their not. Factoring by grouping doesn't work either.

A third step is to see if a convenient substitution can work. You've probably been asked to factor a quartic like x4+x2−6 in the past. By setting y=x2 , this expression becomes y2+y−6, and now your task has become a lot simpler. But for the quartic in your problem, this algebra 2 method doesn't work either. You must be taking precalculus!

In precalculus, you're expected to know certain facts about polynomials. Every precalculus course requires that you know the Fundamental Theorem of Algebra. If you include multiplicity then that fourth degree polynomial will have four roots. If you were looking to solve for x by setting that polynomial equal to zero, this theorem would let you know to expect four solutions at most. You should know how the roots of a polynomial set equal to zero relate to the factors of the polynomial, and that relationship leads to another polynomial theorem.

According to the Linear Factors Theorem, you should be able to factor that thing all the way to the form (x−c1)(x−c2)(x−c3)(x−c4) where the four constants are complex numbers. Those four constants might or might not be real, and they might or might not be repeated more than once. This theorem provides some guidance for you with this problem. If we completely factor that quartic, we'll have a maximum of four linear terms.

Descartes' Rule of Signs, the Conjugate Roots Theorem for Complex Roots, the Conjugate Roots Theorem for Irrational Roots, and theorems about upper and lower limits for roots also might help you, and your course might expect you to know most or all of these. Descartes' Rule of Signs can be usefully applied to your expression, but we're going to use the Rational Roots Theorem.

According to the Rational Roots Theorem, for polynomials with integer coefficients, every rational root of a polynomial, p/q, must have p as a factor of the constant term of the polynomial and q as a factor of the leading coefficient of the polynomial. This means you should be able to factor a term (x−p/q) out of the polynomial. You can understand why the Rational Roots Theorem works by applying it to quadratics. We'll start out by knowing the factored form of a quadratic equation with two rational roots:

(x−p1/q1)(x−p2/q2)=0

Multiplying both sides by q1q2 gives us:

q1(x−p1/q1)q2(x−p2/q2)=0 or

(q1x−p1)(q2x−p2)=0. After expanding by using FOIL, we get:

q1q2x2−(p1q1+p2q2)x+ p1p2=0.

Notice how the x2 coefficient has factors that are the denominators of the rational roots and the units coefficient has factors that are the numerators.

Of course, the rational roots theorem only works for polynomials with rational roots. But if you're given a quartic in a precalculus course that can't be factored by another method, you can bet that it will have rational roots. If it didn't, you wouldn't be able to answer the question without using a calculator or computer.

For your problem, we're looking for roots of the form p/q where p is a factor of −8 and q is a factor of 1. The possibilities are ±1, ±2, ±4, and ±8. Test these possibilities one at a time, starting with the easiest numbers, 1 and −1. You can just plug these numbers into the expression and see if it evaluates to zero. However, there can be advantages to using synthetic division while testing rational roots because you can apply theorems about the upper and lower limits of roots to gain information about whether an incorrect attempt was too large or too small. Descartes's Rule of Signs can help you narrow down the possibilities, but let's move on the fact that −1 will work!

After you find this first root, you'll need to use synthetic division to determine the quotient when your original polynomial is divided by the factor corresponding to that root.

(x4−3x3+4x2−8)/(x+1) = x3−4x2+8x−8.

Congratulations! Now instead of factoring a quartic, you only need to factor a cubic. Try using the same theorems and tests again. Here's what your final answer should be:

http://www.wolframalpha.com/input/?i=Factor+x^4-3x^3%2B4x^2-8

Thank you very much for your great response Mr Marc,A great teacher. Highly recommend you for students.

Nish

Marc, I can only commend your thoroughness!

Luke B. | fun + learning = inevitable progressfun + learning = inevitable progress
4.9 4.9 (90 lesson ratings) (90)
1

One thing you can do is graph is and your factors will be where the y-value equals zero (where the graph crosses the x-axis).

You could also factor it as: x^2(x^2 - 3x + 4) - 8

then to: x^2( x ( x - 3 ) + 4) - 8

set equal to zero: x^2( x ( x - 3 ) + 4) - 8 = 0

move the eight over: x^2( x ( x - 3 ) + 4) = 8

and determine that -1 is one root, and 2 is the other