Question

factor $-4t^{4}-t^{2}+5$ completely.

Explanation:

Step1: Rearrange the polynomial

First, we can rewrite the polynomial in standard form (descending powers of \( t \)): \( -4t^4 - t^2 + 5=-(4t^4 + t^2 - 5) \)

Step2: Let \( u = t^2 \), then the quadratic in \( u \) is \( 4u^2+u - 5 \)

We need to factor \( 4u^2 + u - 5 \). We look for two numbers that multiply to \( 4\times(-5)=-20 \) and add up to \( 1 \). The numbers are \( 5 \) and \( -4 \).

So we can rewrite the middle term:
\( 4u^2+5u - 4u - 5 \)

Step3: Factor by grouping

Group the first two terms and the last two terms:
\( (4u^2 + 5u)+(-4u - 5)=u(4u + 5)-1(4u + 5)=(4u + 5)(u - 1) \)

Step4: Substitute back \( u = t^2 \)

We get \( (4t^2 + 5)(t^2 - 1) \)

Step5: Factor \( t^2 - 1 \) (difference of squares)

\( t^2 - 1=(t + 1)(t - 1) \)

Step6: Remember the negative sign from Step1

Putting it all together, the original polynomial factors as:
\( - (4t^2 + 5)(t + 1)(t - 1) \) or we can also write it as \( -(t - 1)(t + 1)(4t^2 + 5) \)

Answer:

\(-(t - 1)(t + 1)(4t^2 + 5)\) (or equivalent factored form like \(-(4t^2 + 5)(t - 1)(t + 1)\))