Question

Using the Remainder and Factor Theorems, factorise the polynomial. \(x^3 + 10x^2 - 37x + 26.\) 

[Hint : Let \(p(x) = x^3 + 10x^2 - 37x + 26\). Then, \(p(1) = 0\).
\(\therefore (x - 1)\) is a factor of \(p(x)\).]

Answer 1

Let \(p(x) = x^{3} + 10x^{2} - 37x + 26\).

By Factor Theorem, if \(p(1) = 0\), then \((x - 1)\) is a factor. 

\(p(1) = 1 + 10 - 37 + 26 = 0\).

So, \((x - 1)\) is a factor. 

Dividing \(p(x)\) by \((x - 1)\), we get quotient \(x^{2} + 11x - 26\). 

Therefore, \(p(x) = (x - 1)(x^{2} + 11x - 26)\). 

Now factorise the quadratic: \(x^{2} + 11x - 26 = (x + 13)(x - 2)\).

Hence, \(x^{3} + 10x^{2} - 37x + 26 = (x - 1)(x - 2)(x + 13)\).