how do you show that x + 1 is a factor of x^19 + x^12?

Question

and how do you solve an equation like (4-x)^3 = 64 + 20x^2 - 9x^3

Answer 1

Easy.

if you want to show that a polynomial can be divided by (x-a) , a must be a root of the polynomial

In x^19 +x^12 if you put x= -1 you get 0 (-1)^19 =1 , (-1) ^12 = +1

so -1 is a root and the polynomial can be divided by (x+1)

Answer 2

(1) Use factor theorem which states (for this expression) that if (x + 1) is a factor of x^19 + x^12 then when x = -1 f(x) must = 0.

Thus, for (x + 1) to be a factor (x + 1) = 0 or x = -1. This gives the expression -1^19 + (-1^12) which is -1 +1 which = 0 so it is a factor.

(2)

(a) Multiply out the LHS

(b) Simplify by gathering like terms from both sides. You will end up with a cubic equation.

(c) Find a factor of the expression. You might have to try a few before you get one. Use the factor theorem as per part (1).

(d) Reduce the cubic expression to this factor x quadratic expression.

(e) Solve the quadratic equation as per usual.

(f) You will end up with 3 factors of the cubic expression. Set each of these to = 0 and you will get three values for x. These are the solutions or roots. Some of them could be equal of course and you would have only two solutions.

Answer 3

Two ways to show x+1 is a factor of x^19 + x^12.

[1] If x+1 is a factor then x=-1 is a root of the polynomial ie when x=-1 the value of the whole polynomial is 0.

So (-1)^19 = -1
and (-1)^12 = +1
--------------------------
Total = 0
So x+1 is a root.

[2] Method two
Express x^19 +x^12 as x^12(x^7+1)
You know that -1 raised to any odd power is -1 so clearly x=-1 is a root of x^7+1

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Question 2
Expand the left hand side to
64-48x+12x^2-x^3

Simplify the two side by subtracting common amounts
-48x = 8x^2-8x^3

Simplify again by dividing by common factors
-6 = x - x^2

so x^2 - x - 6 = 0

factoring you get (x-3)(x+2) x=3 or x = -2
(or you might just see that straight away from -6 = x(1-x) )

Answer 4

If r is a root of a polynomial, then (x-r) will exactly divide the polynomial. So let r = -1 and substitute -1 for x in the polynomial to ger (-1)^19 + (-1)^12 = -1 + 1 = 0 so (x+1 will exactly divide x^19 + x^12.

In the 2'nd problem, multiply out the (4-x)^3 term and reduce the expression to something manageable that can be factored.

Good luck ☺


Doug

Answer 5

Remainder theorem: f(-1)=(-1)^19 + (-1)^12 = -1+1 = 0
Hence x+1 must be a factor.

(4-x)(4-x)(4-x) - 64 - 20x^2 + 9x^3 = 0
(4-x)(16-8x+x^2) - 64 - 20x^2 + 9x^3 = 0
64 - 32x + 4x^2 - 16x + 8x^2 - x^3 - 64 - 20x^2 + 9x^3 = 0
8x^3 - 8x^2 - 48x = 0
8x (x^2 - x - 6) = 0
8x (x + 2)(x - 3) = 0
x = 0, -2 or 3

Answer 6

Its really easy if you use the elementary Theorem of Factors.
If the equation is represented by P(x), then x+1 is a factor if
p(-1)=0. [ as x+1=0 or x=-1]
Now,
p(-1)= (-1)^19 + (-1)12
= -1 + 1
= 0

Answer 7

f(x)=x^19+x^12
f(-1)=(-1)^19+(-1)^12
=-1+1
=0
sox-1 is a factor

2,.(4-x)^3=-9x^3+20x^2+64
64-48x+12x^2-x^3=-9x^3+20x^2+64
8x^3-8x^2-48x=0
8x(x^2-8x-6)=0
either 8x=0
or x^2-8x-6=0
solve x^2-8x-6=0
using the quadratic formula

Answer 8

first..let y =x^9+x^12

x+1=0,x=-1

replace x=-1 in y..if it equals 2 zero,x+1 is a factor..

n z 2nd..send it ol 2 z r.h.s,n let it b n equation..do it by trial,using difrnt values of x..until the equation =0..there it is,its solved..

Answer 9

x+1=0
x=-1
then
(-1)^19+(-1)^12 =-1+1=0
The value zero, denotes x+1 is a factor.

To solve the equation
collect all the terms to one side.
Then use horner method

Answer 10

1. = x^12[1 + x^7]
= x^12[1 + x -x +x^7]
=x^12[(1+ x) -x(1-x^6)]
=x^12[(1+x)-x(1+x^3)(1-x^3)]
=x^12[(1+x)-x(1-x^3)(1 +x -x+ x^3)]
=x^12[(1+x) - x(1-x^3)({1+x} - x{1-x^2})]
=x^12 [(1+x) -x(1-x^3)({1+x} -x{(1+x)(1-x)})]
= x^12[(1+x) -x(1-x^3)(1+x)(1 -x(1-x))
=x^12(1+x)[1 - x(1-x^3)(1 - x(1-x))]
hence proved

2. use (a-b)^3 = a^3 -b^3 - 3ba^2 + 3ab^2 -6ab