Question

show that 3x+5 is a factor of 3x^3-16x^2-5x+50​ pls answer right and very fast

Answer 1

Answer:

Use synthetic division to determine whether x – 4 is a factor of:

–2x5 + 6x4 + 10x3 – 6x2 – 9x + 4

For x – 4 to be a factor, you must have x = 4 as a zero. Using this information, I'll do the synthetic division with x = 4 as the test zero on the left:

completed division

Since the remainder is zero, then x = 4 is indeed a zero of –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4, so:

Yes, x – 4 is a factor of –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4

Find all the factors of 15x4 + x3 – 52x2 + 20x + 16 by using synthetic division.

Remember that, if x = a is a zero, then x – a is a factor. So use the Rational Roots Test (and maybe a quick graph) to find a good value to test for a zero (x-intercept). I'll try x = 1:

completed division

This division gives a zero remainder, so x = 1 must be a zero, which means that x – 1 is a factor. Since I divided a linear factor (namely, x – 1) out of the original polynomial, then my result has to be a cubic: 15x3 + 16x2 – 36x – 16. So I need to find another zero before I can apply the Quadratic Formula. I'll try x = –2:

Step-by-step explanation:

Answer 2

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

Show that 3x + 5 is a factor of 3x³ - 16x² - 5x + 50.

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

$3x + 5 = 0$

$3x = - 5$

$x = - \frac{5}{3}$

Put the value of x in 3x³ - 16x² - 5x + 50.

$p(x) = 3x {}^{3} - 16 {x}^{2} - 5x + 50$

$p( - \frac{5}{3} ) = 3( - \frac{5}{3} ) {}^{3} - 16 {( - \frac{5}{3} )}^{2} - 5( - \frac{5}{3} ) + 50$

$p( - \frac{5}{3} ) = 3 \times - \frac{125}{27} - 16 \times \frac{25}{9} - 5 \times - \frac{5}{3} + 50$

$p( - \frac{5}{3} ) = - \frac{125}{9} - \frac{400}{9} + \frac{25}{3} + 50$

$p( - \frac{5}{3} ) = - \frac{175}{3} + \frac{25}{3} + 50$

$p( - \frac{5}{3} ) = - 50 + 50$

$p( - \frac{5}{3} ) = 0$

Hence, the value of p(-5/3) = 0

$\implies$ So, we can say that 3x + 5 is a factor of 3x³ - 16x² - 5x + 50.

$\large \textbf \purple {Hence \:PROVED!!}$

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