Question

(a⁴-19a²+9)/(a²-5a+3)= ???

Answer 1

`(a^4-19a^2+9)/(a^2-5a+3)=color(red)(a^2+5a+3)`

Explanation:

Using long polynomial division:
#{: (,ul(color(white)("x")),ul(a^2),ul(+),ul(5a),ul(+),ul(3),ul(color(white)("x")),ul(color(white)("XX")),ul(color(white)("x")),ul(color(white)("X"))), (a^2-5a+3,")",a^4,+,0a^3,-,19a^2,+,0a,+,9), (,,ul(a^4),ul(-),ul(5a^3),ul(+),ul(3a^2),,,,), (,,,,5a^3,-,22a^2,+,0a,,), (,,,,ul(5a^3),ul(-),ul(25a^2),ul(+),ul(15a),,), (,,,,,,3a^2,-,15a,+,9), (,,,,,,ul(3a^2),ul(-),ul(15a),ul(+),ul(9)), (,,,,,,,,,,0) :}#

Answer 2

`a^2+5a+3.`

Explanation:

In `a^4-19a^2+9,` we split `-19a^2` as `6a^2-25a^2.`

Note that, `6a^2` is the Middle Term required to complete

the Square `a^4+9.`

`:. a^4-19a^2+9=a^4+6a^2+9-25a^2,`

`=(a^2+3)^2-(5a)^2,`

`=(a^2+5a+3)(a^2-5a+3).`

`rArr (a^4-19a^2+9)/(a^2-5a+3)=a^2+5a+3.`

Answer 3

`a^2+5a+3`

Explanation:

`"one way is to divide out using the divisor as a factor in the"`
`"numerator"`

`"consider the numerator"`

`color(red)(a^2)(a^2-5a+3)color(magenta)(+5a^3-3a^2)-19a^2+9`

`=color(red)(a^2)(a^2-5a+3)color(red)(+5a)(a^2-5a+3)color(magenta)(+25a^2cancel(-15a))cancel(+9)`

`=color(red)(a^2)(a^2-5a+3)color(red)(+5a)(a^2-5a+3)`

`color(white)(=)color(red)(+3)(a^2-5a+3)color(magenta)(cancel(+15a-9))`

`color(red)(a^2)(a^2-5a+3)color(red)(+5a)(a^2-5a+3)color(red)(+3)(a^2-5a+3)+0`

`rArr(cancel((a^2-5a+3))(a^2+5a+3))/cancel((a^2-5a+3))=a^2+5a+3`

Answer 4

You can find the quotient through long division, or the following nifty trick!

Explanation:

In much the same way that `a/b=c` is the same as `bxxc=a`, if a polynomial `P(a)` satisfies `(a^4-19a^2+9)/(a^2-5a+3)=P(a)`, that same polynomial also satisfies:

`(a^2-5a+3)xxP(a)=(a^4-19a^2+9)`

This `P(a)` is the polynomial we wish to find.

Here's the trick:

(Note: this trick is not meant to require more pen-and-paper work. The following work is meant to illustrate the method; you should be able to do the arithmetic in your head.)

If such a `P(a)` exists, its first term needs to multiply by `a^2` to create the `a^4`. Thus, we must begin our polynomial with `a^4/a^2`, or `a^2`, and the rest of the terms will have lower degree:

`(a^2-5a+3)xx(a^2+color(green)Ba+color(green)C)=(a^4-19a^2+9)`

Because of this, the new `a^2` in `P(a)` will get multiplied by the `-5a` to create a `(–5a^3)` term on the LHS. We don't want this; there is `0a^3` on the right. We need to introduce a term in `P(a)` that cancels this off.

The terms on the LHS that create `a^3` terms on the right are the `(a^2xxBa)` pair and the `(–5axxa^2)` pair. So we want `(a^2xxBa)+(–5axxa^2)` on the LHS to equal `0a^3` on the RHS. This can easily be solved for `B`:

`color(white)=>Ba^3-5a^3=0a^3`
`=>Ba^3"           "=5a^3" "=>" "B=5`

Our equation is now:

`(a^2-5a+3)xx(a^2+5a+color(green)C)=(a^4-19a^2+9)`

We use the same method to find `C`. The pairs on the LHS that create `a^2` terms are `(a^2xxC)`, `(–5axx5a)`, and `(3xxa^2)`. These need to sum to the `–19a^2` on the RHS:

`(a^2xxC)+(–5axx5a)+(3xxa^2)=–19a^2`
`"    "Ca^2"     "-"       "25a^2"    "+"     "3a^2"     "=–19a^2`
`"    "Ca^2"                                                  "="     "3a^2`

Thus, `C=3`.

Our equation is now:

`(a^2-5a+3)xx(a^2+5a+3)=(a^4-19a^2+9)`

We should double-check to make sure this completed LHS produces the RHS exactly. The pairs on the LHS that make `a` terms are `–5axx3` and `3xx5a`. The sum of these is indeed `0a`, which matches the RHS.

The only pair on the LHS that makes a constant term is `3xx3`, which matches the `9` on the RHS.

So that's it! We've found `P(a)` to be `(a^2+5a+3)`.