Anita claims that for any value of c, the number x = 1 is a zero of p(x) = x^2 - (c+1)x + c. Check her claim by substituting x = 1. Is x = 1 always a zero?

Question

Accepted Answer

Yes, for every value of c. p(1) = 1 - (c+1) + c = 1 - c - 1 + c = 0. The c terms cancel, so x = 1 is always a zero.

Anita claims that for any value of $c$ , the number $x = 1$ is a zero of $p (x) = x^{2} - (c + 1) x + c$ .
Check her claim by substituting $x = 1$ .
Is $x = 1$ always a zero?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Zeroes of a Polynomial practice

Anita claims that for any value of c, the number x=1 is a zero of p(x)=x2−(c+1)x+c.Check her claim by substituting x=1.Is x=1 always a zero?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Zeroes of a Polynomial practice

Anita claims that for any value of $c$ , the number $x = 1$ is a zero of $p (x) = x^{2} - (c + 1) x + c$ .
Check her claim by substituting $x = 1$ .
Is $x = 1$ always a zero?