factorise :
x^4 - 15x^2 + 1
Answers
Answer:
Equation at the end of step 1
((x4) - (3•5x2)) - 16 = 0
STEP
2
:
Trying to factor by splitting the middle term
2.1 Factoring x4-15x2-16
The first term is, x4 its coefficient is 1 .
The middle term is, -15x2 its coefficient is -15 .
The last term, "the constant", is -16
Step-1 : Multiply the coefficient of the first term by the constant 1 • -16 = -16
Step-2 : Find two factors of -16 whose sum equals the coefficient of the middle term, which is -15 .
-16 + 1 = -15 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -16 and 1
x4 - 16x2 + 1x2 - 16
Step-4 : Add up the first 2 terms, pulling out like factors :
x2 • (x2-16)
Add up the last 2 terms, pulling out common factors :
1 • (x2-16)
Step-5 : Add up the four terms of step 4 :
(x2+1) • (x2-16)
Which is the desired factorization
Polynomial Roots Calculator :
2.2 Find roots (zeroes) of : F(x) = x2+1
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is 1.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 2.00
1 1 1.00 2.00
Polynomial Roots Calculator found no rational roots
Trying to factor as a Difference of Squares:
2.3 Factoring: x2-16
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 16 is the square of 4
Check : x2 is the square of x1
Factorization is : (x + 4) • (x - 4)
Equation at the end of step
2
:
(x2 + 1) • (x + 4) • (x - 4) = 0
STEP
3
:
Theory - Roots of a product
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation:
3.2 Solve : x2+1 = 0
Subtract 1 from both sides of the equation :
x2 = -1
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -1
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 1.0000 i
x= 0.0000 - 1.0000 i
Solving a Single Variable Equation:
3.3 Solve : x+4 = 0
Subtract 4 from both sides of the equation :
x = -4
Solving a Single Variable Equation:
3.4 Solve : x-4 = 0
Add 4 to both sides of the equation :
x = 4
Supplement : Solving Quadratic Equation Directly
Solving x4-15x2-16 = 0 directly
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Solving a Single Variable Equation:
Equations which are reducible to quadratic :
4.1 Solve x4-15x2-16 = 0
This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x2 transforms the equation into :
w2-15w-16 = 0
Solving this new equation using the quadratic formula we get two real solutions :
16.0000 or -1.0000
Now that we know the value(s) of w , we can calculate x since x is √ w
Doing just this we discover that the solutions of
x4-15x2-16 = 0
are either :
x =√16.000 = 4.00000 or :
x =√16.000 = -4.00000 or :
x =√-1.000 = 0.0 + 1.00000 i or :
x =√-1.000 = 0.0 - 1.00000 i