0 added
0 removed
Original
2026-01-01
Modified
2026-02-28
1
<p>To factorize a<a></a><a>cubic polynomial</a>: Find zero using the trail- and-error method. Then, using<a>synthetic division</a>method, divide the given polynomial f(x) by the given binomial \((x-a)\), </p>
1
<p>To factorize a<a></a><a>cubic polynomial</a>: Find zero using the trail- and-error method. Then, using<a>synthetic division</a>method, divide the given polynomial f(x) by the given binomial \((x-a)\), </p>
2
<p>After division, if the remainder is not zero, then \((x-a)\) is not a factor of \(f(x)\).</p>
2
<p>After division, if the remainder is not zero, then \((x-a)\) is not a factor of \(f(x)\).</p>
3
<p>If the remainder is zero, use the division algorithm and write the given polynomial as a<a>product</a>of \((x-a)\) and quadratic quotient \(q(x)\); \(f(x) = (x-a)q(y) + r\)</p>
3
<p>If the remainder is zero, use the division algorithm and write the given polynomial as a<a>product</a>of \((x-a)\) and quadratic quotient \(q(x)\); \(f(x) = (x-a)q(y) + r\)</p>
4
<p>If possible, factor the quadratic quotient further</p>
4
<p>If possible, factor the quadratic quotient further</p>
5
<p>Then, represent the polynomial in<a>factored form</a>.</p>
5
<p>Then, represent the polynomial in<a>factored form</a>.</p>
6
<p>Let’s factor \(f(x) = x^3 - 6x^2 + 11x - 6\) using the aforementioned procedure </p>
6
<p>Let’s factor \(f(x) = x^3 - 6x^2 + 11x - 6\) using the aforementioned procedure </p>
7
<p>The first step is to find a zero using the hit and try method and dividing the given polynomial Try \(x = 1\), \(f(1) = 1^3 - 6(1)^2 + 11(1) - 6\\ = 1 - 6 + 11 - 6\\ = 0 \)</p>
7
<p>The first step is to find a zero using the hit and try method and dividing the given polynomial Try \(x = 1\), \(f(1) = 1^3 - 6(1)^2 + 11(1) - 6\\ = 1 - 6 + 11 - 6\\ = 0 \)</p>
8
<p>So, x = 1 is a zero, a (x-1) is a factor.</p>
8
<p>So, x = 1 is a zero, a (x-1) is a factor.</p>
9
<p>Now, we will use synthetic division to divide f(x) by (x-1) \(\begin{array}{r|rrrr} & 1 & -6 & 11 & -6 \\ 1\, & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array} \)</p>
9
<p>Now, we will use synthetic division to divide f(x) by (x-1) \(\begin{array}{r|rrrr} & 1 & -6 & 11 & -6 \\ 1\, & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array} \)</p>
10
<p> The quotient is: \(x^2 - 5x = 6 \)</p>
10
<p> The quotient is: \(x^2 - 5x = 6 \)</p>
11
<p>Since the remainder is zero, </p>
11
<p>Since the remainder is zero, </p>
12
<p> \(x^2 - 5x + 6 = (x-2) (x-3)\)</p>
12
<p> \(x^2 - 5x + 6 = (x-2) (x-3)\)</p>
13
<p>The polynomial, in its final factored form, is </p>
13
<p>The polynomial, in its final factored form, is </p>
14
<p> \(f(x) = (x-1)(x-2)(x-3)\)</p>
14
<p> \(f(x) = (x-1)(x-2)(x-3)\)</p>
15
<p>Thus, the zeroes of the given polynomial \(x^3 - 6x^2 + 11x - 6 \)are 1, 2, and 3 </p>
15
<p>Thus, the zeroes of the given polynomial \(x^3 - 6x^2 + 11x - 6 \)are 1, 2, and 3 </p>
16
16