# Express as a polynomial. (t^{2} + 2t - 5)(3t^{2} -

Express as a polynomial. $\left({t}^{2}+2t-5\right)\left(3{t}^{2}-t+2\right)$
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Alex Sheppard
Step 1
Given:
$\left({t}^{2}+2t-5\right)\left(3{t}^{2}-t+2\right)$
For the polynomial, we solve the parenthesis,
$⇒{t}^{2}\left(3{t}^{2}-t+2\right)+2t\left(3{t}^{2}-t+2\right)-5\left(3{t}^{2}-t+2\right)$
$⇒3{t}^{4}-{t}^{3}+2{t}^{2}+6{t}^{3}-2{t}^{2}+4t-15{t}^{2}+5t-10$
$⇒3{t}^{4}-5{t}^{3}-15{t}^{2}+9t-10$
Step 2
Hence,
The polynomial is,
$⇒3{t}^{4}-5{t}^{3}-15{t}^{2}+9t-10$
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Bertha Jordan
Consider $\left({t}^{2}+2t-5\right)\left(3{t}^{2}-t+2\right)$
Treating the polynomial $\left(3{t}^{2}-t+2\right)$ as a single real term, and then multiplying, we get
$\left({t}^{2}+2t-5\right)\left(3{t}^{2}-t+2\right)$
$=\left({t}^{2}+2t+\left(-5\right)\right)\left(3{t}^{2}+\left(-t\right)+2\right)$
$={t}^{2}\left(3{t}^{2}+\left(-t\right)+2\right)+2t\left(3{t}^{2}+\left(-t\right)+2\right)+\left(-5\right)\left(3{t}^{2}+\left(-t\right)+2\right)$
[using distributive property]
Again, using distributive property three times, and simplifying the result, we get
$=3{t}^{4}-{t}^{3}+2{t}^{2}+6{t}^{3}-2{t}^{2}+4t-15{t}^{2}+5t-10$
[adding the powers of same base]
$=3{t}^{4}+\left(-1+6\right){t}^{3}+\left(2-2-15\right){t}^{2}+\left(4+5\right)t-10$
[adding the coefficient of like powers of x]
$=3{t}^{4}+5{t}^{3}-15{t}^{2}+9t-10$ [simplifying]
The three monomials in the first polynomial were multiplied by each of the three monomials in the second polynomial, giving us a total of nine terms, whose sum is the required polynomial, is
$3{t}^{4}+5{t}^{3}-15{t}^{2}+9t-10$