Factor Theorem - Deep Thinking

Using the table mode on a calculator allows you to find the roots of a polynomial quickly, and hence factorise it with minimal effort. There's an opportunity here to explore some aspects of polynomials and develop a deeper understanding at A-Level.

Solution

Substituting any positive value for \(x\) will result in the sum of positive values. The only way to achieve \(p(x)=0\) is to have at least one negative term, which comes from cubing a negative value or multiplying a negative value by 20.

Solution

Utilise the fact from the previous question; all roots must be negative because of the positive signs. Also, the linear factors must each have constant terms which multiply to 9, too. That is, if \(p(x)=(x-a)(x-b)(x-c)\) then \(-abc=9\). We either have -1, -1, -9 or -1, -3, -3 as the three roots. In both cases, there is a repeated root.

Solution

Dhriti is using the leading coefficient of 2 to anticipate a factor in the form \((2x+m)\). The assosciated root will be \(x=-\frac{m}{2}\) so using a step of 0.5 will reveal when \(p(x)=0\).

Similarly, for a leading coefficient of 3, Dhriti could use a step of \(\frac{1}{3}\) to spot when \(p(x)=0\).

 

Solution

Hint 1: One bracket must start with \(2x\) and the others are just \(x\)

Hint 3: The signs are all positive so the roots are all negative. The linear brackets will therefore be \((2x+a)(x+b)(x+c)\) where \(a, b, c >0\)

Hint 2: The values \(a\), \(b\) and \(c\) must multiply to 2. They will be 1, 1 and 2 in some order. \(a\neq 2\) because \(2x+2\) has a common factor of 2, meaning all coefficients of \(p(x)\) would be even. \(b\) and \(c\) can be assigned 1 and 2 without loss of generality.

So we get \((2x+1)(x+1)(x+2)\)