Question

consider the following three polynomial f(x)=x²-8x+15 g(x)=2x²-11x+5 h(x)=2x²-7x+3 concider the polynomial p(x)=f(x)g(x)h(x) it turns out that p(x) is a perfect square if square root of p(x)is written as √p(x)=ax³+bx²+cx+d where a is positive the value of 6a+2b+c+d is

Answer 1

Quadratic Polynomials

Given,

f(x)=x²-8x+15

g(x)=2x²-11x+5

h(x)=2x²-7x+3

To find,

P(x)= f(x)g(x)h(x)

=(x²-8x+15)(2x²-11x+5)(2x²-7x+3)

=2x∧4-11x³+5x²-16x³+88x²-40x+30x²-165x+75(2x²-7x+3)

=2x^4-27x³+123x²-205x+75(2x²-7x+3)

=4x^6-14x^5+6x^4-54x^5+189x^4-81x³+246x^4-861x³+369x²-410x³+1435x²-615x+150x²-525x+225

=4x^6-68x^5+441x^4-1352x³+1954x²-1140x+225

∵√P(x)=ax³+bx²+cx+d

= 2x³+x²√(-68x+441)+x√(-1352x+1954)-√1140x+15

∴a=2, b= √(441-68x), c= √(1954-1352x), d=15