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Maths Question 13 – AP-EAMCET 2026

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If a and b are such that x2−x−1 is a factor of ax3+bx2+1, then ab=

If a polynomial Q(x) is a factor of another polynomial P(x), then dividing P(x) by Q(x) should result in a remainder of zero.

Step 1: Understand the Factor Relationship✦ Active

The problem states that x2−x−1 is a factor of ax3+bx2+1. This implies that when ax3+bx2+1 is divided by x2−x−1, the remainder must be zero. We can use the property derived from the factor, which is x2−x−1=0⟹x2=x+1, to simplify the cubic polynomial.

💡 Teacher's Secret Hint

This method is generally more efficient than polynomial long division or finding roots when the factor is a quadratic.

Step 2: Reduce the Cubic Polynomial○ Expand

Substitute x2=x+1 into the polynomial P(x)=ax3+bx2+1 to reduce its degree.

P(x)=ax3+bx2+1 =ax(x2)+bx2+1 =ax(x+1)+b(x+1)+1(Substitute x2=x+1 for the first time) =ax2+ax+bx+b+1 =a(x+1)+ax+bx+b+1(Substitute x2=x+1 for the second time) =ax+a+ax+bx+b+1
💡 Teacher's Secret Hint

Ensure you substitute x2=x+1 for all occurrences of x2 until the highest power of x in the expression is x1.

Step 3: Formulate System of Equations○ Expand

Group the terms by powers of x. For the remainder to be zero, the coefficient of x and the constant term must both be zero.

P(x)=(ax+ax+bx)+(a+b+1) =(2a+b)x+(a+b+1) For P(x) to be divisible by x2−x−1, the remainder must be 0. Thus, we must have: 2a+b=0(Equation 1) a+b+1=0(Equation 2)
Step 4: Solve for a and b○ Expand

Solve the system of linear equations for a and b.

From Equation 1, express b in terms of a: b=−2a Substitute this into Equation 2: a+(−2a)+1=0 −a+1=0 a=1 Now substitute a=1 back into b=−2a: b=−2(1) b=−2
💡 Teacher's Secret Hint

Always check your signs when substituting values to avoid algebraic errors.

Step 5: Calculate ab○ Expand

Finally, calculate the product ab using the determined values of a and b.

ab=(1)(−2) ab=−2
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