Precalculus factoring

Question

Give P(x)=6x^5-17x^4-15x^3+33x^2+13x-12 write P in factored form. Be sure to write the full equation

Answer 1

The factors of -12 are: ±(1,2,3,4,6,12)

P(-1) = 0, so (-1) is a factor

Synthetic division

6 -17 -15 33 13 -12 |-1

-6 23 -8 -25 12

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6 -23 8 25 -12 0 |-1 P₂(-1) = 0

-6 29 -37 12

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6 -29 37 -12 0 |3 P₃(3) = 0

18 -33 12

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6 -11 4

This leaves:

6x²-11x+4 = 0

Doing the slide and divide factoring method

6×4 = 24 Factors of 24 that add to -11 are -8 and -3

(x-8/6)(x-3/6) The divide step

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

P(x)= (x+1)(x+1)(x-3)(3x-4)(2x-1)

Answer 2

(x+1)²(2x-1)(3x-4)(x-3)

Note that all roots are rational. To factor, you can either repeatedly use synthetic division (the rational root theorem gives possibilities, and the factoring above tells you that the roots are -1, -1, 1/2, 4/3, 3 ) or plot the function from -1.5 to 3.2 and see the zero crossings and the double root at -1.

Answer 3

Let find rational zeros: Factors od - 12: ±1, ±2, ±3, ±4, ±6, ±12;

Factors of 6: ±1, ±2, ±3, ±6

List of possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12, ±1/2,±3/2, ±1/3, ±2/3, ±4/3, ±1/6,

Synthetic division gives: x = -1 is zero of multiplicity 2; 1/2, 4/3, 3 are siple zeros.

So, P(x) = 6(x + 1)²(x - 1/2)(x - 4/3)(x - 3)