Question

Is 3x-7 is a factor if polynomial 6x³+x²-26x-21

Answer 1

let,
$g(x) = 3x - 7 \\ \tt \: and \\ f(x) = 6 {x}^{3} + {x}^{2} - 26x - 21$
Now, g ( x ) will be a factor of f ( x ) only if $f( \frac{7}{3} ) = 0$

So, let's ☑ check :

$\quad \: f \left( \frac{7}{3} \right) \\ \\ = 6 { \left( \frac{7}{3} \right)}^{3} + { \left( \frac{7}{3} \right)}^{2} - 26 \left( \frac{7}{3} \right) - 21 \\ \\ = \frac{ {}^{2} \: \cancel{6} \times 343}{9 \: \cancel{27}} + \frac{49}{9} - \frac{26 \times 7}{3} - 21 \\ \\ = \frac{686}{9} + \frac{49}{9} - \frac{182 \times 3}{9} - \frac{21 \times 9}{9} \\ \\ = \frac{686 + 49 - 546 - 189}{9} \\ \\ = \frac{ \cancel{735} - \cancel{735}}{9} \\ \\ = \frac{0}{9} \\ \\ = \bf \: \: 0$

Hence,
$\tt \: (3x - 7) \: is \: \: a \: \: factor \: \: of \: \: (6 {x}^{3} + {x}^{2} - 26x - 21)$

Answer 2

$\color {red} </p><p>\Large \underline {\underline {Question \: :}}$

Is 3x - 7 is a factor of polynomial 6x³ + x² - 26x - 21?

$\color {red} </p><p>\Large \underline {\underline {Answer \: :}}$

Let 3x - 7 and 6x³ + x² - 26x - 21 be g(x) and p(x) respectively.

• g(x) = 3x - 7
• p(x) = 6x³ + x² - 26x - 21

g(x) = 0

3x - 7 = 0

3x = 7

x = $\frac {7}{3}$

Put the value of x in p(x) to find that 3x - 7 is a factor of polynomial 6x³ + x² - 26x - 21 or not.

$p(x) = 6 {x}^{3} + {x}^{2} - 26x - 21$

$p( \frac{7}{3} ) = 6( \frac{7}{3} ) {}^{3} + ( \frac{7}{3} ) {}^{2} - 26( \frac{7}{3} ) - 21$

$p( \frac{7}{3} ) = 6 \times \frac{343}{27} + \frac{49}{9} - 26 \times \frac{7}{3} - 21$

$p( \frac{7}{3} ) = \frac{686}{9} + \frac{49}{9} - \frac{182}{3} - 21$

$p( \frac{7}{3} ) = \frac{686 + 49}{9} - \frac{182}{3} - 21$

$p( \frac{7}{3} ) = \frac{735}{9} - \frac{182}{3} - 21$

$p( \frac{7}{3} ) = \frac{245}{3} - \frac{182}{3} - 21$

$p( \frac{7}{3} ) = \frac{245 - 182}{3} - 21$

$p( \frac{7}{3} ) = \frac{63}{3} - 21$

$p( \frac{7}{3} ) = 21 - 21$

$p( \frac{7}{3} ) = 0$

So, the answer is 0.

$\implies$ Hence, 3x - 7 is a factor of polynomial 6x³ + x² - 26x - 21.