How to solve a quadratic equation that does not factor?

1 answer:

7 0

Use the formula or complete the square.
The zeroes of the quadratic can be real and rational; real and irrational; complex conjugates.
If the quadratic is ax²+bx+c, x=(-b+√b²-4ac)/2a.
If b² > 4ac the solutions are real. If b²-4ac is a perfect square, the solutions are real and rational; otherwise they’re real but irrational.
If b² < 4ac the solutions are complex.

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ruslelena [56]

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3 years ago

Find the value of c so that the polynomial p(x) is divisible by (x-3). p(x) =-x^3+cx^2-4x+3. c=?

Vilka [71]

Answer:

$\boxed{\textsf {The value of c is $\sf \dfrac{-14}{3}$. }}$

Step-by-step explanation:

A polynomial is given to us . We need to find the value of c so that it is divisible by (x-3 ) .The given polynomial to us is ,

$\sf \implies p(x)=- x^3+cx^2-4x+3$

And the factor is ,

$\sf \implies g(x) = x + 3$

Now , according to factor theorem , p(x) will be divisible by g(x) if p(-3) = 0

$\sf \implies p(x)= - x^3+cx^2-4x+3 \\\\\implies\sf p(-3) = 0 \\\\\sf\implies -(-3)^3+c(-3)^2-4(-3)+3 = 0 \\\\\sf\implies 27 + 9c +12+3=0 \\\\\sf\implies 9c +42=0 \\\\\sf\implies 9c = (-42) \\\\\sf\implies c = \dfrac{-42}{9}\\\\\sf\implies \boxed{\pink{\sf c = \dfrac{-14}{3}}}$

<h3>Hence the value of c is (-14)/3 .</h3>

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3 years ago