Question

Find the value of a such that x minus 4 whole bracket is a factor of 5 x cube minus 7 x square minus ax minus 28

Answer 1

Explanation of your question

$f(x) = {5x}^{3} - {7x}^{2} - ax - 28$

x-4=0

x=4

f(x)=0 {because it is remainder}

$0 = 5( {4}^{3} ) - 7 ({4}^{2} ) - 4a - 28$

0=5×64-7×16-4a-28

0=320-112-4a-28

0=208-28-4a

0=180-4a

4a=180

a=45<=Answer

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Answer 2

Answer:

x - 4 = 0

x = 4

p(4) = 5x³ - 7x² - ax - 28

p(4) = 5(4)³ - 7(4)² - a(4) - 28 = 0

p(4) = 320 - 112 - 4a - 28 = 0

p(4) = 4a - 180 = 0

p(4) = 4a = 180

p(4) = a = 180/4

p(4) = a = 45