Question

1) Solve the polynomial equation 4x^4=5x^2+9 by factoring
A: {3/2, i}
B: {3/2, -3/2. i, -1}
C:{3/2.-3/2,1,-1}
2)Simplify the expression (x-y)(x^2+xy+y^2)
A: x^3-y^3
B:x^3-2xy+y^3
C:x^3+y^3
3)A family is traveling by car on a trip and they set their odometer on the car to 0 miles before they left home. After 2 hours in the car, they had traveled 80 miles. When they stopped after 10 hours of tavel, they had driven a total of 496 miles, according to the odometer reading. What is their average rate of speed for their car trip that day?
A: 52mph
B:60mph
C:56mph
Please help!!! thank you!

Answer 1

well start by rewriting the equation

\[4x^4 - 5x^2 - 9 = 0\]

this is an a variation on the quadratic....

so start by multiplying the coefficient of the leading term and the constant

4 * -9 = -36

find the factors of -36 that add to -5, the larger factor is negative...

then you need

(4x^2 + factor 1)(4x^2 + factor 2)
-------------------------------- = 0
4

Answer 2

I think it is A. I am not quite sure. I am trying to understand!!

Answer 3

I don't think that polynomial can be solved by factoring, use the quadratic formula to solve it!

Answer 4

well the problem can be solved by factoring.... and you choice of A is it a guess of from some factoring...

Answer 5

so if you followed what I posted above...

the factors of -36 and -9 and 4

so you are now looking at

\[\frac{(4x^2 + 4)(4x^2 - 9)}{4}\]

the 1st quadratic factor has a common factor of 4

\[\frac{4(x^2 +1)(4x^2 - 9)}{4}\]

remove the common factor of 4 from the numerator and denominator
which gives

\[(x^2 + 1)(4x^2 - 9) = 0\]

so now you are solving 2 quadratics

\[x^2 + 1 = 0....and......4x^2 - 9 = 0\]

solving thre above quadratics will give the solution to the problem

Answer 6

I got a simpler solution!!! let \[ x^{4} = u^{2}, x^{2}=u \] we rewrite the polynomial in terms of u instead of x so it becomes \[4u^{2}-5u-9 \]. Factoring this polynomial gives us \[(4u-9)(u+1)\]. So \[u=\frac{ 9 }{ 4 }, u=-1\]. Remembering that we need x values not u values we write u as \[x^{2} \]. \[x^{2}=\frac{ 9 }{ 4 }, -1 \]. Solving for x gives us \[\frac{ 3 }{ 2 } \] and i.

Answer 7

lol..@Isaiah.Feynman

funny thing ... you said it couldn't be factorised...

Answer 8

I didn't look at it properly at first.