Help with easy factoring problems?

Question

1. Can you solve by factoring? x^4+22x^2-15=0 

2. Solve by division (2x^3-3x+4)÷(x-1) 

3. F(x)=3x^3-11x^2-6x+8 x=4, find the zeros 

4. Write a polynomial function that has real coefficients given the zeroes and leading coefficient of x=5,2,2i

Answer 1

1).

The equation is x⁴ + 22x² - 15 = 0.

Let, x² = t.

x⁴ + 22x² - 15 = 0

(x²)² + 22x² - 15 = 0

t² +22t - 15 = 0.

t² + 22t - 15 = 0., is a quadratic, use quadratic formula to find the roots of the related quadratic equation.

The solution, t =[ - b ± √(b² - 4ac) ]/2a.

Compare the equation with standard form of the quadratic equation ax² + bx + c = 0.

a = 1, b = 22, and c = - 15.

t = [ - 22 ± √(22² - 4 * 1 * (- 15)) ]/2 * 1

t = [ - 22 ± √(484 + 60) ]/2

t = - 11 ± 2√34.

Put t = x² .

x² = - 11 ± 2√34

 x = ± [ - 11 ± 2√34 ]

x = ± [ √( - 11 + 2√34) ] and x = ± [ i√(11 + 2√34) ].

Therefore, the solutions are x = ± [ √( - 11 + 2√34) ] and x = ± [ i√(11 + 2√34) ].

Answer 2

2).

Divident is 2x^3 - 3x + 4.

Divisor is x - 1.

By doing long division,

x - 1 ) 2x^3 - 3x + 4 (2x^2 - 2x + 1

          2x^3 - 2x^2

     (-)______________________

                     2x^2 - 3x + 4

                   - 2x^2 + 2x

                 (+)_______________________

                                - x + 4

                                   x - 1

                        (+)_______________________

                                          3

                            _____________________

Therefore, quotient is 2x^2 - 2x + 1, and remainder is 3.

Answer 3

3).

The function f(x) = 3x³ - 11x² - 6x + 8x = 4.

Rewrite the function as f ( x ) = 3x³ - 11x² + 2x - 4 = 0.

Graph :

The points where the curve crosses the x - axis will give solution to the equation .

The graph crosses the x  - axis at a point that would suggest a solution of the equation.

It crosses the x  - axis at one point , hence there are one real root.

x  = 3.5845

Use synthatic division to detrmine if the given value of x  is a root of the polynomial.

Since f (3.5845) = 0, x = 3.5845 is a zero.

The depressed polynomial is 

Since, the depressed polynomial of this zero, , is a quadratic,

use the Quadratic Formula to find the roots of the related quadratic equation

Compare the depressed polynomial with standard form of quadratic equation .

a = 3, b = - 0.2465, and c = 1.1164.

Substitute the values a = 3, b = - 0.2465, and c = 1.1164.

.

The zeros of the function are .

Answer 4

4).

Note :

If the function has a complex root, then it's complex conjugate is also a root.

since complex roots only occur as complex conjugate pairs.

If the zeroes of the polynomail function are x = 5, 2, and 2i,

Then the polynomial function is

f(x) = (x - 5) * (x - 2) * (x - 2i) * (x + 2i)

      = (x^2 - 5x - 2x + 10)(x^2 + 4)

      = (x^2 - 7x + 10)(x^2 + 4)

      = x^4 - 7x^3 + 14x^2 - 28x + 40.

Therefore, the polynomial function is x^4 - 7x^3 + 14x^2 - 28x + 40.