What must be added to $5 x^{3} - 2 x^{2} + 6 x + 7$ to make the sum $x^{3} + 3 x^{2} - x + 1$

Open in App

Solution

Lets $y$ is to be added $5 x^{3} - 2 x^{2} + 6 x + 7$ to make the sum $x^{3} + 3 x^{2} - x + 1$
$\Rightarrow y = (x^{3} + 3 x^{2} - x + 1) - (5 x^{3} - 2 x^{2} + 6 x + 7)$
$= x^{3} + 3 x^{2} - x + 1 - 5 x^{3} + 2 x^{2} - 6 x - 7$
$= x^{3} - 5 x^{3} + 3 x^{2} + 2 x^{2} - x - 6 x - 7 + 1$
$= - 4 x^{3} + 5 x^{2} - 7 x - 6$
Thus, $- 4 x^{3} + 5 x^{2} - 7 x - 6$ should be added to $5 x^{3} - 2 x^{2} + 6 x + 7$ to get $x^{3} + 3 x^{2} - x + 1$

Suggest Corrections

Similar questions

(3 x^{3} - 2 x^{2} + 3 x - 1)

(x^{3} + 2 x^{2} - 6 x + 5) ?

2 x^{2} y^{2} - 5 x + 7

- 5 x^{2} y^{2} + 7 x - 9

What must be subtracted from $x^{3} - 3 x^{2} + 5 x - 1$ to get $x^{3} + 2 x^{2} - 2 x + 11$ ?

What must be added to $2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3$ so that the result is exactly divisible by (x- 2)?

Join BYJU'S Learning Program