For the polynomial \(p(x)=x^3-4x^2+mx+6\), what value of \(m\) will make \(p(1)=0\)?
Answer and explanation
Correct answer: -3
Substituting \(x=1\) gives \(p(1)=1^3-4(1)^2+m(1)+6=1-4+m+6=m+3\). Since \(p(1)=0\), we require \(m+3=0\), so \(m=-3\). The nearby option \(-2\) would give \(p(1)=1\), not zero. Exam tip: when a value is given as a zero of a polynomial, substitute it and set the resulting expression equal to zero.
Frequently asked questions
What is the correct answer to this question?
-3
Why is this the correct answer?
Substituting \(x=1\) gives \(p(1)=1^3-4(1)^2+m(1)+6=1-4+m+6=m+3\). Since \(p(1)=0\), we require \(m+3=0\), so \(m=-3\). The nearby option \(-2\) would give \(p(1)=1\), not zero. Exam tip: when a value is given as a zero of a polynomial, substitute it and set the resulting expression equal to zero.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Polynomials in one variable.
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