Express as a polynomial. (t2+2t−5)(3t2−t+2)

Question

Alex Sheppard · Accepted Answer

Step 1  Given:  (t2+2t−5)(3t2−t+2)  For the polynomial, we solve the parenthesis,  ⇒t2(3t2−t+2)+2t(3t2−t+2)−5(3t2−t+2)  ⇒3t4−t3+2t2+6t3−2t2+4t−15t2+5t−10  ⇒3t4−5t3−15t2+9t−10  Step 2  Hence,  The polynomial is,  ⇒3t4−5t3−15t2+9t−10

Bertha Jordan · Accepted Answer

Consider (t2+2t−5)(3t2−t+2)  Treating the polynomial (3t2−t+2) as a single real term, and then multiplying, we get  (t2+2t−5)(3t2−t+2)  =(t2+2t+(−5))(3t2+(−t)+2)  =t2(3t2+(−t)+2)+2t(3t2+(−t)+2)+(−5)(3t2+(−t)+2)  [using distributive property]  Again, using distributive property three times, and simplifying the result, we get  =3t4−t3+2t2+6t3−2t2+4t−15t2+5t−10  [adding the powers of same base]  =3t4+(−1+6)t3+(2−2−15)t2+(4+5)t−10  [adding the coefficient of like powers of x]  =3t4+5t3−15t2+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  3t4+5t3−15t2+9t−10

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

Answered question

Answer & Explanation

New Questions in Algebra II