Subtract the second polynomial from the first and write the degree of the polynomial so obtained.

(i) x⁴ + x² + x $-$ 1 ; x⁴ $-$ x³ $-$ x² $-$ 1
(ii) n³ $-$ 5n² + 6 ; n² $-$ 3n + 8
(iii) 2a + 3a² $-$ 7 ; 3a² $-$ 12 + 2a

Open in App

Solution

$(i) Required polynomial = (x^{4} + x^{2} + x - 1) - (x^{4} - x^{3} - x^{2} + 1) = x^{4} + x^{2} + x - 1 - x^{4} + x^{3} + x^{2} - 1 = x^{4} - x^{4} + x^{3} + x^{2} + x^{2} + x - 1 - 1 = x^{3} + 2 x^{2} + x - 2 Here, the highest power is 3 . ∴ Degree of the polynomial = 3$

$(ii) Required polynomial = (n^{3} - 5 n^{2} + 6) - (n^{3} - 3 n + 8) = n^{3} - 5 n^{2} + 6 - n^{3} + 3 n - 8 = n^{3} - n^{3} - 5 n^{2} + 3 n + 6 - 8 = - 5 n^{2} + 3 n - 2 Here, the highest power is 2 . ∴ Degree of the polynomial = 2$

$(iii) Required polynomial = (2 a + 3 a^{2} - 7) - (3 a^{2} - 12 + 2 a) = 2 a + 3 a^{2} - 7 - 3 a^{2} + 12 - 2 a = 3 a^{2} - 3 a^{2} + 2 a - 2 a - 7 + 12 = 5 Here, the polynomial is a nonzero constant . W e know that the degree of a nonzero constant is 0 . ∴ Degree of the polynomial = 0$

Suggest Corrections

Similar questions

2 a + 3 a^{2} - 7; 3 a^{2} - 12 + 2 a

x^{4} + x^{2} + x - 1; x^{4} - x^{3} - x^{2} + 1

n^{3} - 5 n^{2} + 6; n^{3} - 3 n + 8

Prove that none of the polynomials listed below has first degree factors:

(i) x² + x + 1

(ii) x⁴ + 1

(iii) x² − x + 1

(iv) x⁴ + x² + 1

The degree of the polynomial obtained when $8 - 6 x + x^{2} - 7 x^{3} + x^{5}$ is subtracted from $x^{4} - 6 x^{3} + x^{2} - 3 x + 1$ is:

Join BYJU'S Learning Program