how do you factor the polynomial x^4+8x^2-9?

x^4+8x^2-9 = x^4 + 9x^2 - x^2 - 9
= x^2 (x^2 + 9) - 1 (x^2 + 9)
= (x^2 - 1) (x^2 + 9)
= (x + 1) (x - 1) (x^2 + 9)

x^4 + 8x^2 - 9
= x^4 + 9x^2 - x^2 - 9
= (x^4 + 9x^2) - (x^2 + 9)
= x^2(x^2 + 9) - 1(x^2 + 9)
= (x^2 + 9)(x^2 - 1)
= (x^2 + 9)(x^2 - 1^2)
= (x^2 + 9)(x + 1)(x - 1)

x^4 + 8x - 9

Factor a trinomial

(x^2 - 1) (x^2 + 9)

Factor a difference between two squares

(x + 1) (x - 1) (x^2 + 9)

Answer:

(x + 1) (x - 1) ( x^2 + 9)

(x^2+9)(x^2-1)=

(x^2+9)(x+1)(x-1)=

( x ² + 9 ) ( x ² - 1 ) = ( x ² + 9 ) ( x - 1 ) ( x + 1 )

Rearranging the polynomial; x^4 + 8x^2 - 9 = x^4 + 9x^2 -x^2 - 9

=> x^2 (x^2 + 9) - 1(x^2 + 1)

=> (x^2 - 1)(x^2 + 9)

=> (x - 1)(x + 1)(x^2 + 9) (Ans)

you can further factorise but factors will be imaginary.

=> (x - 1)(x + 1)(x^2 - √(-9)^2]

=> (x - 1)(x + 1)(x + 3i)(x - 3i)

where i = √(-1), an imaginary number.

Same way you would factor a quadratic ax^2 + bx + c = 0.

So if you have x^4 + 8x^2 - 9 = 0, you would first factor it so that:

(x^2 + 9) * (x^2 - 1) = 0. Then look at each factored term individually.

x^2 + 9 cannot be factored anymore, but x^2 - 1 can be factored. So now we have:

(x^2 + 9) * (x + 1) * (x - 1) = 0.

Now just set each factored term equal to zero and solve.

You end up with x = -1 and x = 1.

Same way as you factor ax^2+bx-c, except in this case it's degree of 4 and 2, just think what two numbers adds to 8 and multiplys to -9 and you are good. Think about it before you scroll down to look at the answer.
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(x^2+9)(x^2-1)
= (x^2+9)(x+1)(x-1)

(x^2+9) (x^2-1)

x^4+8x^2-9

=x^4+9x^2-x^2-9 [by middle term factorization]

=x^2(x^2+9)-(x^2+9)

=(x^2+9)*(x^2-1) [as a^2-b^2=(a+b)(a-b)]
[and i^2= -1]
={(x)^2-(3i)^2}*(x-1)*(x+1)[i=the imaginary number,positive square root of -1,i.e i=sqrt(-1)]
=(x+1)*(x-1)*(x+3i)*(x-3i)
(answer)